<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

327. Yet Another Palindrome
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



You are given N words. Find the shortest palindrome (a word that is the same with its reverse) containing all of them as contiguous substrings.

Input
The first line of input contains an integer N, 1 ≤ N ≤ 14. The next N lines contain one word each, between 1 and 30 characters long. All the characters are small English letters ('a' through 'z').

Output
Output the required palindrome. If there're several possible solutions, output any.

Example(s)
sample input
sample output
1
avtobus
avtobusubotva

sample input
sample output
3
bacd
edcab
cabac

<|response|>
## 1) Abridged problem statement

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

---

## 2) Key observations needed to solve the problem

1. **Redundant words can be removed.**  
   If a word `w` (or `reverse(w)`) occurs as a substring of a *longer* word `u` (or `reverse(u)`), then any solution containing `u` automatically contains `w`. So `w` can be discarded.

2. **A palindrome is fully determined by its “right half”.**  
   If we construct some string `S` that will be the right half, then the full palindrome is:
   - even center: `reverse(S) + S`
   - odd center: `reverse(S)[1:] + S` (one shared middle character)

3. **Every word may be placed in either orientation.**  
   Because the whole result is a palindrome, if a substring appears on one side, its reverse appears mirrored on the other. Thus it’s natural to allow each word to be used as either `w` or `reverse(w)` when building the half.

4. **We want a shortest common superstring (SCS)-like chain on the half.**  
   If we chain oriented words with maximum suffix→prefix overlaps, the half becomes shorter.  
   Adding `x` characters to the half usually adds `2x` characters to the final palindrome.

5. **The first word determines the palindrome center savings.**  
   If the starting oriented word has a **palindromic prefix**, part of it can be shared around the center (especially for an odd-length center), reducing the total length.

---

## 3) Full solution approach

### Step A — Read, deduplicate, and filter redundant words
1. Read all words.
2. Remove duplicates.
3. Sort by length.
4. For each word `w`, if it appears as a substring inside any longer word `u` in **any orientation** (`u`, `rev(u)`) and for `w` also (`w`, `rev(w)`), discard `w`.
5. Keep remaining words and store both orientations:
   - `words[i][0] = w`
   - `words[i][1] = reverse(w)`

After this pruning, we still have `n ≤ 14`.

---

### Step B — Build oriented “nodes” and overlaps
There are `2n` oriented nodes:
- node `2*i` = `words[i][0]`
- node `2*i+1` = `words[i][1]`

Precompute overlap:
`overlap[a][b]` = maximum `k` such that **suffix** of string `a` of length `k` equals **prefix** of string `b` of length `k`, where `a` and `b` come from **different original words**.

---

### Step C — DP over subsets (SCS-style, but cost doubled)
We build the **right half** as an ordered chain of oriented nodes.

State:
- `dp[mask][c]` = minimal **final palindrome length** when we have used the set of original words `mask`, and the last oriented node is `c` (`0..2n-1`).

Transitions:
- From `(mask, c)`, add a new word `j` not in `mask`, choose orientation `rev`:
  - `nc = 2*j + rev`
  - new characters added to half = `len(word[nc]) - overlap[c][nc]`
  - palindrome grows by `2 * (len - overlap)`
  - update `dp[mask | (1<<j)][nc]`

Initialization (choosing the first node and the center):
- For each start node `c`:
  - Let `w` be its string.
  - Compute `best_pal` = length of **longest palindromic prefix** of `w`.
  - We can skip `best_pal//2` characters into `w` when defining the initial half seed:
    - `half = w[best_pal//2:]`
  - If `best_pal` is odd, we have an **odd center** (one shared character).
  - Initial palindrome length:
    - `2*len(half)` if even center
    - `2*len(half) - 1` if odd center
  - Set `dp[1<<i][c]` accordingly and store `half` + `odd/even` for reconstruction.

---

### Step D — Reconstruct best chain and output palindrome
1. Choose `best_c` minimizing `dp[full_mask][c]`.
2. Backtrack with `prev_state` to recover the path of oriented nodes.
3. Build the right-half string by concatenating with overlaps.
4. Mirror it to produce the palindrome:
   - even center: `reverse(half) + half`
   - odd center: `reverse(half)[1:] + half` (equivalently drop 1 char from the mirrored side)

This palindrome contains all filtered words, therefore contains all original input words.

---

## 4) C++ implementation with detailed comments

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

/*
  327. Yet Another Palindrome

  Approach:
  - Remove duplicates.
  - Filter out words that are substrings of some longer word in any orientation.
  - For remaining n (<=14), consider each word in 2 orientations => 2n nodes.
  - Precompute overlaps between oriented nodes.
  - DP over subsets to build a shortest "right half" chain (SCS-like),
    but every added character to the half costs 2 in final palindrome length.
  - Special initialization for the first word:
      use its longest palindromic prefix to decide if we can have an odd center
      and how much of the start is "shared" by mirroring.
  - Reconstruct chain -> build half -> mirror -> output palindrome.
*/

static string revs(const string& s) {
    return string(s.rbegin(), s.rend());
}

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

    int nInput;
    cin >> nInput;
    vector<string> all(nInput);
    for (int i = 0; i < nInput; i++) cin >> all[i];

    // 1) Remove duplicates
    sort(all.begin(), all.end());
    all.erase(unique(all.begin(), all.end()), all.end());

    // 2) Sort by length so we can test "contained in longer word"
    sort(all.begin(), all.end(), [](const string& a, const string& b) {
        return a.size() < b.size();
    });

    // 3) Filter redundant words
    vector<array<string,2>> words; // words[i][0]=original, words[i][1]=reversed
    for (int i = 0; i < (int)all.size(); i++) {
        string w = all[i];
        string rw = revs(w);
        bool keep = true;

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

            string u = all[j];
            string ru = revs(u);

            // If w or rw occurs in u or ru, then w is redundant
            if (u.find(w) != string::npos || u.find(rw) != string::npos ||
                ru.find(w) != string::npos || ru.find(rw) != string::npos) {
                keep = false;
                break;
            }
        }

        if (keep) {
            words.push_back({w, rw});
        }
    }

    int n = (int)words.size();
    int m = 2 * n;                // number of oriented nodes
    int FULL = (1 << n) - 1;

    // Edge case: if after filtering there's nothing (shouldn't happen), but safe:
    if (n == 0) {
        cout << "\n";
        return 0;
    }

    // 4) Precompute overlaps overlap[a][b]
    vector<vector<int>> overlap(m, vector<int>(m, 0));
    for (int a = 0; a < m; a++) {
        const string& A = words[a / 2][a % 2];
        for (int b = 0; b < m; b++) {
            if (a / 2 == b / 2) continue; // don't chain same original word
            const string& B = words[b / 2][b % 2];

            int lim = min(A.size(), B.size());
            int best = 0;
            for (int k = 1; k <= lim; k++) {
                // check if suffix of A with length k equals prefix of B with length k
                if (A.compare((int)A.size() - k, k, B, 0, k) == 0) best = k;
            }
            overlap[a][b] = best;
        }
    }

    auto longestPalPrefix = [&](const string& s) -> int {
        int best = 1;
        for (int len = 1; len <= (int)s.size(); len++) {
            bool ok = true;
            for (int l = 0, r = len - 1; l < r; l++, r--) {
                if (s[l] != s[r]) { ok = false; break; }
            }
            if (ok) best = len;
        }
        return best;
    };

    const int INF = 1e9;

    // dp[mask][c] = minimal final palindrome length
    vector<vector<int>> dp(1 << n, vector<int>(m, INF));
    // prev node used for backtracking; prevState[mask][c] = previous oriented node
    vector<vector<int>> prevState(1 << n, vector<int>(m, -1));

    // For reconstruction of the half: starting half seed and odd/even center choice
    vector<string> startHalf(m);
    vector<char> startOdd(m, 0);

    // 5) Initialize DP by choosing each word as the "central starter" in either orientation
    for (int i = 0; i < n; i++) {
        for (int r = 0; r < 2; r++) {
            int c = 2 * i + r;
            const string& w = words[i][r];

            int bestPal = longestPalPrefix(w);
            int halfStart = bestPal / 2;          // shared by mirroring
            string half = w.substr(halfStart);    // right-half seed
            bool odd = (bestPal % 2 == 1);

            int palLen = 2 * (int)half.size() - (odd ? 1 : 0);

            int mask = 1 << i;
            dp[mask][c] = palLen;
            startHalf[c] = half;
            startOdd[c] = odd ? 1 : 0;
        }
    }

    // 6) Subset DP transitions
    for (int mask = 1; mask <= FULL; mask++) {
        for (int c = 0; c < m; c++) {
            if (dp[mask][c] >= INF) continue;

            for (int j = 0; j < n; j++) {
                if (mask & (1 << j)) continue; // not used yet?

                for (int r = 0; r < 2; r++) {
                    int nc = 2 * j + r;
                    int ov = overlap[c][nc];
                    int addHalf = (int)words[j][r].size() - ov;

                    // every added char in half adds 2 chars in final palindrome
                    int cost = 2 * addHalf;

                    int nmask = mask | (1 << j);
                    int nd = dp[mask][c] + cost;
                    if (nd < dp[nmask][nc]) {
                        dp[nmask][nc] = nd;
                        prevState[nmask][nc] = c;
                    }
                }
            }
        }
    }

    // 7) Choose best ending node
    int bestC = 0;
    for (int c = 1; c < m; c++) {
        if (dp[FULL][c] < dp[FULL][bestC]) bestC = c;
    }

    // 8) Reconstruct path of oriented nodes
    vector<int> path;
    {
        int mask = FULL;
        int c = bestC;
        while (true) {
            path.push_back(c);
            int pc = prevState[mask][c];
            if (pc == -1) break;
            mask &= ~(1 << (c / 2)); // remove the word corresponding to current node
            c = pc;
        }
        reverse(path.begin(), path.end());
    }

    // 9) Build right-half string using overlaps
    string half = startHalf[path[0]];
    for (int idx = 1; idx < (int)path.size(); idx++) {
        int prev = path[idx - 1];
        int cur  = path[idx];
        int ov = overlap[prev][cur];
        const string& s = words[cur / 2][cur % 2];
        half += s.substr(ov);
    }

    // 10) Mirror half to get full palindrome
    string left(half.rbegin(), half.rend());
    string ans;
    if (startOdd[path[0]]) {
        // odd center: middle character shared => drop 1 char from mirrored side
        left.pop_back();
        ans = left + half;
    } else {
        ans = left + half;
    }

    cout << ans << "\n";
    return 0;
}
```

---

## 5) Python implementation with detailed comments

```python
import sys

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

def longest_pal_prefix(s: str) -> int:
    """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 a[-k:] == 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() -> None:
    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)]

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

    # 2) Filter redundant words:
    # drop w if it occurs in any longer u in any orientation (u or reverse(u)),
    # and also consider w or reverse(w).
    kept = []
    for i, w in enumerate(all_words):
        rw = reverse(w)
        keep = 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):
                keep = False
                break
        if keep:
            kept.append(w)

    # Store both orientations
    words = [(w, reverse(w)) for w in kept]
    n = len(words)
    if n == 0:
        print()
        return

    m = 2 * n
    FULL = (1 << n) - 1

    # 3) 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)

    INF = 10**9
    # dp[mask][c] = minimal final palindrome length
    dp = [[INF] * m for _ in range(1 << n)]
    prev = [[-1] * m for _ in range(1 << n)]

    # Start data for reconstruction
    start_half = [""] * m
    start_odd = [False] * m

    # 4) Initialization: choose first word/orientation as the center starter
    for i in range(n):
        for r in (0, 1):
            c = 2 * i + r
            w = words[i][r]

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

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

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

    # 5) Subset DP transitions
    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 r in (0, 1):
                    nc = 2 * j + r
                    ov = overlap[c][nc]
                    add_half = len(words[j][r]) - ov
                    cost = 2 * add_half

                    nmask = mask | (1 << j)
                    nd = cur + cost
                    if nd < dp[nmask][nc]:
                        dp[nmask][nc] = nd
                        prev[nmask][nc] = c

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

    # 7) Backtrack path of oriented nodes
    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 the set
        c = pc
    path.reverse()

    # 8) Build right half by applying overlaps
    half = start_half[path[0]]
    for idx in range(1, len(path)):
        p = path[idx - 1]
        q = path[idx]
        ov = overlap[p][q]
        half += words[q // 2][q % 2][ov:]

    # 9) Mirror to build palindrome
    left = half[::-1]
    if start_odd[path[0]]:
        left = left[:-1]  # shared center char
    ans = left + half
    print(ans)

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

