## 1) Abridged problem statement

Given **N (1 ≤ N ≤ 14)** lowercase words (length ≤ 30), output **any shortest palindrome string** that contains **each given word as a contiguous substring**.  

---

## 2) Detailed editorial (how the provided solution works)

### Key observations

1. **If a word appears as a substring of another word (or of its reverse), it’s redundant.**  
   Suppose word `x` is contained in some longer word `y` (in some orientation: `y`, `reverse(y)`), and we ensure the final palindrome contains `y`. Then the final palindrome automatically contains `x` too.  
   So we can safely remove such redundant words to reduce the problem size.

2. **A palindrome is determined by one “half”.**  
   If we build the right half string `S`, then the palindrome is:
   - even center: `reverse(S) + S`
   - odd center: `reverse(S without last char) + S` (the middle char is shared)

3. **Handling words and their reverses.**  
   In a palindrome, if a substring appears somewhere, its reverse appears symmetrically on the other side. This motivates allowing each word to be placed in **either orientation** when we build our half-chain.

4. **We want to build a superstring with maximal overlaps.**  
   Like shortest common superstring (SCS), we want to order chosen oriented strings to maximize overlaps between consecutive strings.  
   But because we’re building a palindrome from a half, **each extra character added on the half contributes 2 characters** in the final palindrome (one on each side), except possibly the central shared character in the odd case.

---

### Step A: Filter redundant words

- Read all words.
- Remove duplicates.
- For each word `w`, check if it is a substring of **any longer** word `u` in **any orientation**:
  - `u` contains `w`
  - `u` contains `reverse(w)`
  - `reverse(u)` contains `w`
  - `reverse(u)` contains `reverse(w)`
- If yes, drop `w`. Otherwise keep it.
- Store each kept word as a pair: `{w, reverse(w)}`.
- After filtering, the effective `n` is small (still ≤ 14 in worst case).

This pruning is essential for speed and for avoiding useless DP states.

---

### Step B: Precompute overlaps between all oriented forms

We have `2*n` oriented “nodes”:
- node `2*i` = word `i` in original orientation
- node `2*i+1` = word `i` reversed

For each ordered pair `(a, b)` with `a` and `b` from **different original words**, compute:

`max_overlap[a][b]` = maximum `k` such that suffix of `a` of length `k` equals prefix of `b` of length `k`.

This is the standard overlap used in superstring chaining.

---

### Step C: DP over subsets to choose the best chain (SCS-style)

State:
- `dp[mask][c]` = minimal palindrome length achievable after covering the set `mask` of words, where the **last appended oriented node** is `c` (`0..2n-1`).

We also store `prev_state[mask][c]` to reconstruct the chain.

#### Initialization: choosing the first word and the palindrome center

The first chosen oriented word `w` sits around the center of the palindrome.  
We compute the **longest palindromic prefix** of `w`, call its length `best_pal`.

Why? Because if the prefix is already palindromic, part of it can lie on both sides with less duplication. Concretely:

- Let `half_start = best_pal // 2`
- Let `half = w[half_start:]`  (this is what we keep as the “right-half seed”)
- If `best_pal` is odd, there is a shared center character (odd palindrome).

If `half` has length `L`:
- even center: initial palindrome length is `2*L`
- odd center: initial palindrome length is `2*L - 1`

So:
- `dp[1<<i][c] = initial_length`
- record `start_str[c] = half`
- record `start_odd[c] = (best_pal % 2 == 1)`

#### Transition: append a new word (in either orientation)

From state `(mask, c)` append word `j` not in `mask`, and choose orientation `rev`:

- `nc = 2*j + rev`
- overlap = `max_overlap[c][nc]`
- adding `words[j][rev].size() - overlap` new chars to the half
- palindrome grows by **twice** that:  
  `cost = 2 * (len(new_word) - overlap)`

Update:
`dp[mask | (1<<j)][nc] = min(dp[mask][c] + cost)`

This is essentially SCS DP but with doubled costs.

---

### Step D: Reconstruct the best chain and build the answer

- Pick `best_c` minimizing `dp[full_mask][c]`.
- Backtrack using `prev_state` to recover the sequence of oriented nodes `path`.

Build the right-half string `suf`:
- Start with `suf = start_str[path[0]]`
- For each next node, append only the non-overlapping suffix:
  `suf += word(next)[overlap:]`

Finally mirror:
- `rev_suf = reverse(suf)`
- If odd center: `ans = rev_suf[:-1] + suf`
- Else: `ans = rev_suf + suf`

This palindrome contains every chosen word (hence every original word after pruning) as a substring.

Complexity:
- `n ≤ 14`, `2n ≤ 28`
- DP: `O(2^n * (2n) * n * 2)` ≈ feasible.
- Overlap computation: `O((2n)^2 * L^2)` with small limits (L ≤ 30).

---

## 3) Provided C++ solution with detailed line-by-line comments

```cpp
#include <bits/stdc++.h>
using namespace std;

// Pretty-print a pair (mostly unused in this solution, but harmless utility)
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Read a pair
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Read a vector of items
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x : a) in >> x;
    return in;
};

// Print a vector of items (unused)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x : a) out << x << ' ';
    return out;
};

int n;  // number of effective (non-redundant) words after filtering
vector<array<string, 2>> words; // for each i: words[i][0]=original, words[i][1]=reversed

// Reads input, removes duplicates, and filters out redundant words.
void read() {
    cin >> n;
    vector<string> all(n);
    for (int i = 0; i < n; i++) cin >> all[i];

    // Remove duplicates
    sort(all.begin(), all.end());
    all.erase(unique(all.begin(), all.end()), all.end());

    // Sort by length ascending (helps when checking substring-in-longer)
    sort(all.begin(), all.end(), [](const string& a, const string& b) {
        return a.size() < b.size();
    });

    n = 0;          // will recount after filtering
    words.clear();  // ensure empty in case of multiple tests

    // For each word all[i], check if it is contained in any longer word in any orientation
    for (int i = 0; i < (int)all.size(); i++) {
        string rev_i(all[i].rbegin(), all[i].rend()); // reverse(all[i])
        bool ok = true;

        for (int j = 0; j < (int)all.size(); j++) {
            // only compare against strictly longer words
            if (i == j || all[j].size() <= all[i].size()) continue;

            string rev_j(all[j].rbegin(), all[j].rend());

            // If all[i] or its reverse appears in all[j] or reverse(all[j]),
            // then all[i] is redundant.
            if (all[j].find(all[i]) != string::npos ||
                all[j].find(rev_i) != string::npos ||
                rev_j.find(all[i]) != string::npos ||
                rev_j.find(rev_i) != string::npos) {
                ok = false;
                break;
            }
        }

        // Keep only non-redundant words, store both orientations
        if (ok) {
            words.push_back({all[i], rev_i});
            n++;
        }
    }
}

vector<vector<int>> max_overlap;

// Precompute maximum suffix-prefix overlaps between all oriented nodes.
void precompute_transitions() {
    // total oriented nodes = 2*n
    max_overlap.assign(2 * n, vector<int>(2 * n, 0));

    for (int a = 0; a < 2 * n; a++) {
        for (int b = 0; b < 2 * n; b++) {
            // Do not allow using the same original word twice in a row
            if (a / 2 == b / 2) continue;

            // Pick the concrete strings for these oriented nodes
            const string& wa = words[a / 2][a % 2];
            const string& wb = words[b / 2][b % 2];

            // Compute largest k where suffix(wa,k) == prefix(wb,k)
            int lim = min(wa.size(), wb.size());
            int best = 0;
            for (int overlap = 1; overlap <= lim; overlap++) {
                bool ok = true;
                for (int k = 0; k < overlap; k++) {
                    if (wa[wa.size() - overlap + k] != wb[k]) {
                        ok = false;
                        break;
                    }
                }
                if (ok) best = overlap;
            }
            max_overlap[a][b] = best;
        }
    }
}

vector<vector<int>> dp, prev_state;

void solve() {
    /*
      We build the palindrome by constructing its "right half" as a chain
      of chosen oriented words with maximal overlaps.

      dp[mask][c] = minimal palindrome length when we already included
      words in 'mask' and the last oriented node used is 'c'.

      Initialization chooses the first oriented word; its longest palindromic
      prefix allows saving characters around the center (odd/even center).
    */

    precompute_transitions();

    int full = (1 << n) - 1;

    // Large value for infinity
    dp.assign(1 << n, vector<int>(2 * n, (int)1e9));
    prev_state.assign(1 << n, vector<int>(2 * n, -1));

    // For reconstructing the final half:
    // start_str[c] is the initial half string when starting with oriented node c
    vector<string> start_str(2 * n);
    // start_odd[c] indicates whether the palindrome center is odd for this start
    vector<bool> start_odd(2 * n, false);

    // Initialize DP for each word i in each orientation rev
    for (int i = 0; i < n; i++) {
        for (int rev = 0; rev < 2; rev++) {
            int c = 2 * i + rev;                 // oriented node index
            int len_i = words[i][rev].size();

            // Find the longest palindromic prefix length of words[i][rev]
            int best_pal = 1;
            for (int len = 1; len <= len_i; len++) {
                bool is_pal = true;
                for (int a = 0, b = len - 1; a < b; a++, b--) {
                    if (words[i][rev][a] != words[i][rev][b]) {
                        is_pal = false;
                        break;
                    }
                }
                if (is_pal) best_pal = len;
            }

            // If prefix length best_pal is palindromic, then in the half
            // we can "skip" best_pal/2 characters because they will be mirrored.
            int half_start = best_pal / 2;

            // The starting right-half string seed
            string half = words[i][rev].substr(half_start);
            int half_len = (int)half.size();

            // Odd center if best_pal is odd
            bool odd_center = (best_pal % 2 == 1);

            // The palindrome length produced by mirroring this half.
            int pal_len = 2 * half_len - (odd_center ? 1 : 0);

            // Mask includes only word i
            int init_mask = 1 << i;

            dp[init_mask][c] = pal_len;
            start_str[c] = half;
            start_odd[c] = odd_center;
        }
    }

    // Subset DP: try appending a not-yet-used word in either orientation
    for (int mask = 1; mask <= full; mask++) {
        for (int c = 0; c < 2 * n; c++) {
            if (dp[mask][c] >= (int)1e8) continue; // unreachable

            for (int j = 0; j < n; j++) {
                if (mask & (1 << j)) continue; // already used word j

                for (int rev = 0; rev < 2; rev++) {
                    int nc = 2 * j + rev;              // next oriented node
                    int overlap = max_overlap[c][nc];  // overlap on the half

                    // Additional characters added to the half = len - overlap
                    // Palindrome grows by twice that
                    int cost = 2 * ((int)words[j][rev].size() - overlap);

                    int new_mask = mask | (1 << j);
                    int new_len = dp[mask][c] + cost;

                    if (new_len < dp[new_mask][nc]) {
                        dp[new_mask][nc] = new_len;
                        prev_state[new_mask][nc] = c; // remember predecessor node
                    }
                }
            }
        }
    }

    // Choose best ending oriented node
    int best_c = 0;
    for (int c = 1; c < 2 * n; c++) {
        if (dp[full][c] < dp[full][best_c]) best_c = c;
    }

    // Reconstruct path of oriented nodes by backtracking prev_state
    vector<int> path;
    {
        int mask = full;
        int c = best_c;

        while (prev_state[mask][c] != -1) {
            path.push_back(c);
            int prev_c = prev_state[mask][c];

            // Remove the word corresponding to node c from the mask
            mask &= ~(1 << (c / 2));
            c = prev_c;
        }
        path.push_back(c);
        reverse(path.begin(), path.end());
    }

    // Build the right-half string 'suf' by concatenating with overlaps
    string suf = start_str[path[0]];
    for (int i = 1; i < (int)path.size(); i++) {
        int c = path[i];
        int prev_c = path[i - 1];
        int overlap = max_overlap[prev_c][c];

        // Append only the non-overlapping part of the next word
        suf += words[c / 2][c % 2].substr(overlap);
    }

    // Mirror to build the full palindrome
    string rev_suf(suf.rbegin(), suf.rend());
    string ans;

    // If odd center, the middle character is shared, so drop 1 char from left mirror
    if (start_odd[path[0]]) {
        ans = rev_suf.substr(0, rev_suf.size() - 1) + suf;
    } else {
        ans = rev_suf + suf;
    }

    cout << ans << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for (int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

## 4) Python solution (same idea) with detailed comments

```python
import sys

def reverse(s: str) -> str:
    return s[::-1]

def longest_pal_prefix(s: str) -> int:
    """Return length of the longest prefix of s that is a palindrome."""
    best = 1
    for L in range(1, len(s) + 1):
        pref = s[:L]
        if pref == pref[::-1]:
            best = L
    return best

def max_suffix_prefix_overlap(a: str, b: str) -> int:
    """Maximum k such that suffix(a,k) == prefix(b,k)."""
    lim = min(len(a), len(b))
    best = 0
    for k in range(1, lim + 1):
        if a[-k:] == b[:k]:
            best = k
    return best

def solve():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    n_in = int(next(it))
    all_words = [next(it) for _ in range(n_in)]

    # Remove duplicates
    all_words = sorted(set(all_words), key=len)

    # Filter redundant words: drop w if it appears in any longer word in any orientation
    kept = []
    for i, w in enumerate(all_words):
        rw = reverse(w)
        ok = True
        for j, u in enumerate(all_words):
            if i == j or len(u) <= len(w):
                continue
            ru = reverse(u)
            if (w in u) or (rw in u) or (w in ru) or (rw in ru):
                ok = False
                break
        if ok:
            kept.append(w)

    # Build words[i][0]=original, words[i][1]=reversed
    words = [(w, reverse(w)) for w in kept]
    n = len(words)
    m = 2 * n  # number of oriented nodes

    # Precompute overlaps between oriented nodes
    overlap = [[0] * m for _ in range(m)]
    for a in range(m):
        wa = words[a // 2][a % 2]
        for b in range(m):
            if a // 2 == b // 2:
                continue
            wb = words[b // 2][b % 2]
            overlap[a][b] = max_suffix_prefix_overlap(wa, wb)

    FULL = (1 << n) - 1
    INF = 10**9

    # dp[mask][c] = minimal palindrome length
    dp = [[INF] * m for _ in range(1 << n)]
    prev = [[-1] * m for _ in range(1 << n)]

    # start_half[c] and start_odd[c] allow reconstructing final palindrome
    start_half = [""] * m
    start_odd = [False] * m

    # Initialize: choose first word i with orientation rev
    for i in range(n):
        for rev in (0, 1):
            c = 2 * i + rev
            w = words[i][rev]

            best_pal = longest_pal_prefix(w)
            half_start = best_pal // 2
            half = w[half_start:]   # right-half seed
            odd_center = (best_pal % 2 == 1)

            pal_len = 2 * len(half) - (1 if odd_center else 0)

            mask = 1 << i
            dp[mask][c] = pal_len
            start_half[c] = half
            start_odd[c] = odd_center

    # DP transitions: append a new word j (either orientation)
    for mask in range(1, FULL + 1):
        for c in range(m):
            if dp[mask][c] >= INF:
                continue
            for j in range(n):
                if mask & (1 << j):
                    continue
                for rev in (0, 1):
                    nc = 2 * j + rev
                    ov = overlap[c][nc]
                    # cost added to palindrome length is double the new chars on the half
                    cost = 2 * (len(words[j][rev]) - ov)
                    nmask = mask | (1 << j)
                    new_len = dp[mask][c] + cost
                    if new_len < dp[nmask][nc]:
                        dp[nmask][nc] = new_len
                        prev[nmask][nc] = c

    # Find best ending node
    best_c = min(range(m), key=lambda c: dp[FULL][c])

    # Reconstruct oriented node path
    path = []
    mask = FULL
    c = best_c
    while prev[mask][c] != -1:
        path.append(c)
        mask ^= (1 << (c // 2))  # remove that word from the set
        c = prev[mask | (1 << (path[-1] // 2))][path[-1]]  # not used; ignore
    # The line above is messy; do it cleanly with a loop storing prev first.

def solve_clean():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    n_in = int(next(it))
    all_words = [next(it) for _ in range(n_in)]

    all_words = sorted(set(all_words), key=len)

    kept = []
    for i, w in enumerate(all_words):
        rw = w[::-1]
        ok = True
        for j, u in enumerate(all_words):
            if i == j or len(u) <= len(w):
                continue
            ru = u[::-1]
            if (w in u) or (rw in u) or (w in ru) or (rw in ru):
                ok = False
                break
        if ok:
            kept.append(w)

    words = [(w, w[::-1]) for w in kept]
    n = len(words)
    m = 2 * n

    overlap = [[0] * m for _ in range(m)]
    for a in range(m):
        wa = words[a // 2][a % 2]
        for b in range(m):
            if a // 2 == b // 2:
                continue
            wb = words[b // 2][b % 2]
            lim = min(len(wa), len(wb))
            best = 0
            for k in range(1, lim + 1):
                if wa[-k:] == wb[:k]:
                    best = k
            overlap[a][b] = best

    def longest_pal_prefix(s: str) -> int:
        best = 1
        for L in range(1, len(s) + 1):
            pref = s[:L]
            if pref == pref[::-1]:
                best = L
        return best

    FULL = (1 << n) - 1
    INF = 10**9
    dp = [[INF] * m for _ in range(1 << n)]
    prev = [[-1] * m for _ in range(1 << n)]

    start_half = [""] * m
    start_odd = [False] * m

    for i in range(n):
        for rev in (0, 1):
            c = 2 * i + rev
            w = words[i][rev]
            best_pal = longest_pal_prefix(w)
            half = w[best_pal // 2:]
            odd = (best_pal % 2 == 1)
            dp[1 << i][c] = 2 * len(half) - (1 if odd else 0)
            start_half[c] = half
            start_odd[c] = odd

    for mask in range(1, FULL + 1):
        for c in range(m):
            cur = dp[mask][c]
            if cur >= INF:
                continue
            for j in range(n):
                if mask & (1 << j):
                    continue
                for rev in (0, 1):
                    nc = 2 * j + rev
                    ov = overlap[c][nc]
                    cost = 2 * (len(words[j][rev]) - ov)
                    nmask = mask | (1 << j)
                    nl = cur + cost
                    if nl < dp[nmask][nc]:
                        dp[nmask][nc] = nl
                        prev[nmask][nc] = c

    best_c = min(range(m), key=lambda c: dp[FULL][c])

    # Backtrack
    path = []
    mask = FULL
    c = best_c
    while True:
        path.append(c)
        pc = prev[mask][c]
        if pc == -1:
            break
        mask ^= (1 << (c // 2))  # remove current word from mask
        c = pc
    path.reverse()

    # Build right half
    suf = start_half[path[0]]
    for i in range(1, len(path)):
        prev_c = path[i - 1]
        c = path[i]
        ov = overlap[prev_c][c]
        suf += words[c // 2][c % 2][ov:]

    rev_suf = suf[::-1]
    if start_odd[path[0]]:
        ans = rev_suf[:-1] + suf
    else:
        ans = rev_suf + suf

    sys.stdout.write(ans + "\n")

if __name__ == "__main__":
    solve_clean()
```

---

## 5) Compressed editorial

- Remove duplicates; discard any word that (in either orientation) is a substring of a longer word in either orientation.  
- For remaining `n ≤ 14`, treat each word in two orientations (original/reverse) → `2n` nodes.
- Precompute `overlap[a][b]`: max suffix-prefix match between oriented strings `a → b` (disallow same original word).
- DP over subsets like shortest common superstring:
  - `dp[mask][c]` = minimal palindrome length covering `mask`, last node `c`.
  - Initialize each start node using the longest palindromic prefix to decide odd/even center and initial half.
  - Transition by appending a new word with maximal overlap; adding `k` chars to half adds `2k` to palindrome length.
- Reconstruct best path, build right-half string by overlaps, mirror it (drop 1 char on mirror if odd center) to output the palindrome.