## 1) Concise abridged problem statement

You are given two strings of uppercase letters:

- First tape string `A` of length `N`
- Second tape string `B` of length `M`, with `M < N`

A **cyclic shift** moves characters left/right with wrap-around (e.g., left shift `"ABRA"` → `"BRAA"`).

To open the lock:

1. Cyclically shift `B` to obtain the **lexicographically smallest** cyclic shift of `B`. Call this resulting string `B'`.
2. Cyclically shift `A` so that its first `M` characters match `B'` **with at most one mismatching position**.

Output:
- Line 1: `B'`
- Line 2: the chosen shifted version of `A`

If multiple `A`-shifts work, choose one with **minimum number of shifts**; if still tied, choose the **left shift** direction.

A solution is guaranteed to exist.

---

## 2) Detailed editorial (explaining the provided solution)

### Key observations

#### A) Handling cyclic shifts by doubling the string
A cyclic shift corresponds to taking a length-`L` substring of `S+S`:
- Any cyclic shift of `S` (length `L`) equals `S2[i : i+L)` where `S2 = S + S` and `0 ≤ i < L`.

So we build:
- `A2 = A + A` (length `2N`)
- `B2 = B + B` (length `2M`)

Then:
- A cyclic shift of `B` is `B2[i : i+M)`, `i in [0, M-1]`
- A cyclic shift of `A` is `A2[i : i+N)`, `i in [0, N-1]`

#### B) Need fast substring comparisons ⇒ rolling hash + LCP by binary search
To compare two candidate cyclic shifts lexicographically, we need to find the first position where they differ (their LCP).

We use Rabin–Karp polynomial hashing over mod \(2^{61}-1\) (a fast “Mersenne-like” prime), which supports:
- `hash(S[l..r])` in O(1) after O(n) preprocessing
- LCP between two substrings by binary searching the length and comparing hashes: **O(log L)**

Define:
`lcp(i, j, max_len)` = longest common prefix length between substrings starting at positions `i` and `j` with maximum length `max_len`.

---

### Step 1: Find lexicographically smallest cyclic shift of `B`

We want:
\[
best\_b = \arg\min_{0 \le i < M} B2[i:i+M)
\]

To compare shifts `i` and `j`:
1. `l = lcp(i, j, M)`
2. If `l == M`, they are equal.
3. Else compare `B2[i+l]` vs `B2[j+l]`.

This is O(M log M) time (M comparisons, each LCP O(log M)).

---

### Step 2: Find a shift of `A` whose prefix matches `B'` with ≤ 1 mismatch

Let `B' = B2[best_b : best_b+M)`.

For each candidate starting position `pos_a` in `A2` (meaning shifted `A` is `A2[pos_a : pos_a+N)`), we need:
The first `M` chars of that shifted `A` differ from `B'` in at most one position.

#### Counting mismatches using at most two LCP calls
We compute mismatches in the length-`M` window:
- Compute `l1 = lcp(pos_a, best_b, M)` between `A2[pos_a:]` and `B2[best_b:]`.
  - If `l1 == M`: exact match → 0 mismatches
  - Otherwise the first mismatch is at index `l1`.
- Skip that mismatching character and match the remainder:
  - Compute `l2 = lcp(pos_a + l1 + 1, best_b + l1 + 1, M - l1 - 1)`
  - If `l1 + 1 + l2 == M`, then everything else matches → exactly 1 mismatch
  - Otherwise → at least 2 mismatches

So mismatch-counting is O(log M).

---

### Tie-breaking on the shift of `A`

We must:
1. Minimize number of shifts needed to reach the chosen cyclic shift from original `A`.
2. If tied, prefer left shift.

If we represent a left shift by `d` positions as taking `A2[d : d+N)`, then:
- Left shift by `d` corresponds to start index `d`
- Right shift by `d` corresponds to left shift by `N-d` (start index `N-d`)

The solution searches in increasing `d = 0..floor(N/2)` and checks:
- left shift `d`
- then (if different) right shift `d` i.e. start at `N-d`

This exactly implements:
- minimal shift count first (smallest `d`)
- and for the same `d`, left shift is checked before right shift

Guaranteed solution exists, so it will find one.

---

### Complexity
- Hash preprocessing: O(N + M)
- Smallest shift of `B`: O(M log M)
- Finding shift of `A`: up to N candidates worst-case, each O(log M) ⇒ O(N log M)

Overall: **O(N log N)**-ish (dominant term O(N log M)), fits constraints with fast hashing.

---

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

```cpp
#include <bits/stdc++.h>
// All standard headers in one (GCC extension).

#pragma GCC optimize("O3,unroll-loops")
// Aggressive optimization and loop unrolling.

#pragma GCC target("avx2")
// Allow vector instructions; may speed up some operations.

using namespace std;

// Pretty-print a pair (not used in core logic, but harmless).
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Read a pair (also not used).
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Read a vector (also not used).
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Print a vector (also not used).
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// Rolling-hash helper using mod = 2^61-1 with fast multiplication.
class HashMeta {
  private:
    // Choose a random base for hashing to reduce collision risk.
    void set_random_base() {
        seed_seq seed{
            (uint32_t)chrono::duration_cast<chrono::nanoseconds>(
                chrono::high_resolution_clock::now().time_since_epoch()
            )
                .count(),
            (uint32_t)random_device()(), (uint32_t)42
        };
        mt19937 rng(seed);
        base = uniform_int_distribution<uint64_t>(0, mod - 1)(rng);
    }

    // Precompute base^i for i=0..n-1, used to extract substring hashes.
    void precompute_base_pow(size_t n) {
        base_pow.resize(n);
        base_pow[0] = 1;
        for(size_t i = 1; i < n; i++) {
            base_pow[i] = mul(base_pow[i - 1], base);
        }
    }

    // Addition modulo (2^61-1), with reduction trick.
    static constexpr uint64_t add(uint64_t a, uint64_t b) {
        a += b + 1;
        a = (a & mod) + (a >> 61);
        return a - 1;
    }

    // Subtraction modulo.
    static constexpr uint64_t sub(uint64_t a, uint64_t b) {
        return add(a, mod - b);
    }

    // Multiplication modulo (2^61-1), using 128-bit-like splitting trick.
    static constexpr uint64_t mul(uint64_t a, uint64_t b) {
        uint64_t l1 = (uint32_t)a, h1 = a >> 32, l2 = (uint32_t)b, h2 = b >> 32;
        uint64_t l = l1 * l2, m = l1 * h2 + l2 * h1, h = h1 * h2;
        uint64_t ret =
            (l & mod) + (l >> 61) + (h << 3) + (m >> 29) + (m << 35 >> 3) + 1;
        ret = (ret & mod) + (ret >> 61);
        ret = (ret & mod) + (ret >> 61);
        return ret - 1;
    }

  public:
    // Wrapper type around uint64_t hash value, with operators.
    class hash_t {
        uint64_t h;

      public:
        hash_t() : h(0) {}
        hash_t(uint64_t h) : h(h) {}
        operator uint64_t() const { return h; }

        hash_t& operator+=(const hash_t& other) {
            h = add(h, other.h);
            return *this;
        }

        hash_t& operator-=(const hash_t& other) {
            h = sub(h, other.h);
            return *this;
        }

        hash_t& operator*=(const hash_t& other) {
            h = mul(h, other.h);
            return *this;
        }

        hash_t operator+(const hash_t& other) const {
            return hash_t(*this) += other;
        }
        hash_t operator-(const hash_t& other) const {
            return hash_t(*this) -= other;
        }
        hash_t operator*(const hash_t& other) const {
            return hash_t(*this) *= other;
        }

        bool operator==(const hash_t& other) const { return h == other.h; }
        bool operator!=(const hash_t& other) const { return h != other.h; }

        // Useful for ordered containers if needed.
        bool operator<(const hash_t& other) const { return h < other.h; }
    };

    uint64_t base;              // random base
    vector<hash_t> base_pow;    // base powers
    static constexpr uint64_t mod = (1ull << 61) - 1;

    // Initialize hashing for strings up to length n (powers computed up to n).
    void init(size_t n) {
        set_random_base();
        precompute_base_pow(n);
    }

    // Build prefix hashes for a container (string) in O(len).
    // h[i] = hash of container[0..i].
    template<typename T>
    vector<hash_t> rabin_karp(const T& container) {
        vector<hash_t> h(container.size());
        for(size_t i = 0; i < container.size(); i++) {
            h[i] = (i ? h[i - 1] : hash_t(0)) * hash_t(base) +
                   hash_t(container[i]); // uses character code as number
        }
        return h;
    }

    // Return hash of substring [l..r] inclusive from prefix hash array h.
    hash_t hash_range(int l, int r, const vector<hash_t>& h) {
        if(l == 0) {
            return h[r];
        }
        return h[r] - h[l - 1] * base_pow[r - l + 1];
    }
};

HashMeta hash_meta;
using hash_t = HashMeta::hash_t;

int n, m;
string A, B;

void read() {
    cin >> n >> m;
    cin >> A >> B;
    // Double both strings so that cyclic shifts become substrings.
    A = A + A;
    B = B + B;
}

void solve() {
    // Prepare hashing; A length is 2n, B length is 2m.
    // base_pow size uses 2*n; that is enough for all hash extractions
    // (we only query substrings within A of length <= n and within B of length <= m).
    hash_meta.init(2 * n);

    // Prefix hashes for doubled strings.
    vector<hash_t> ph_a = hash_meta.rabin_karp(A);
    vector<hash_t> ph_b = hash_meta.rabin_karp(B);

    // Longest Common Prefix between:
    // - substring of first string starting at i
    // - substring of second string starting at j
    // up to max_len characters.
    // Uses binary search on length and hash comparisons.
    auto lcp = [&](int i, int j, vector<hash_t>& ph_i, vector<hash_t>& ph_j,
                   int max_len) {
        int lo = 0, hi = max_len;
        while(lo < hi) {
            int mid = (lo + hi + 1) / 2;
            // Compare hashes of length mid.
            if(hash_meta.hash_range(i, i + mid - 1, ph_i) ==
               hash_meta.hash_range(j, j + mid - 1, ph_j)) {
                lo = mid;      // they match for mid, try bigger
            } else {
                hi = mid - 1;  // mismatch, try smaller
            }
        }
        return lo;
    };

    // Compare two cyclic shifts of B (starting at i and j) lexicographically,
    // each of length m.
    auto cmp_cyclic = [&](int i, int j) {
        int l = lcp(i, j, ph_b, ph_b, m);
        if(l == m) {
            return false; // equal -> not less
        }
        return B[i + l] < B[j + l]; // compare first differing character
    };

    // Find best_b = starting index of lexicographically minimal cyclic shift of B.
    int best_b = 0;
    for(int i = 1; i < m; i++) {
        if(cmp_cyclic(i, best_b)) {
            best_b = i;
        }
    }

    // Count mismatches between:
    // A substring A[pos_a : pos_a+m) and B substring B[best_b : best_b+m),
    // but only needs to distinguish 0/1 vs 2+ mismatches.
    auto count_mismatches = [&](int pos_a) {
        // First LCP: match from the start.
        int l1 = lcp(pos_a, best_b, ph_a, ph_b, m);
        if(l1 == m) {
            return 0; // full match
        }
        if(l1 == m - 1) {
            return 1; // only last character differs
        }

        // Assume mismatch at position l1, skip it in both strings and match the rest.
        int l2 = lcp(pos_a + l1 + 1, best_b + l1 + 1, ph_a, ph_b, m - l1 - 1);

        // If remainder matches completely, total mismatches is exactly 1.
        if(l1 + 1 + l2 == m) {
            return 1;
        }
        return 2; // at least 2 mismatches
    };

    // Find best_a = start index in A (0..n-1) giving minimal shifts.
    // We search increasing distance d from 0:
    // - check left shift by d first (start index d),
    // - then right shift by d (start index n-d).
    int best_a = -1;
    for(int d = 0; d <= n / 2 && best_a == -1; d++) {
        if(count_mismatches(d) <= 1) {
            best_a = d; // smallest shift count; left preferred
        } else if(d > 0 && n - d != d && count_mismatches(n - d) <= 1) {
            best_a = n - d; // right shift by d (same shift count), if left failed
        }
    }

    // Construct output strings:
    // - minimal cyclic shift of B is B[best_b..best_b+m)
    // - shifted A is A[best_a..best_a+n)
    string result_b = B.substr(best_b, m);
    string result_a = A.substr(best_a, n);

    cout << result_b << "\n";
    cout << result_a << "\n";
}

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

    int T = 1;
    // Single test case in this problem.
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

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

```python
import sys
import random
import time

# We'll implement rolling hash with mod = 2^61 - 1 (same modulus idea as C++).
# Python integers are arbitrary precision, so we can implement safe reduction.

MOD = (1 << 61) - 1

def mod_add(a, b):
    x = a + b
    x = (x & MOD) + (x >> 61)
    if x >= MOD:
        x -= MOD
    return x

def mod_sub(a, b):
    return mod_add(a, MOD - b)

def mod_mul(a, b):
    # Multiply then reduce mod (2^61-1) using the identity:
    # x mod (2^61-1) = (x & MOD) + (x >> 61), repeated once or twice.
    x = a * b
    x = (x & MOD) + (x >> 61)
    x = (x & MOD) + (x >> 61)
    if x >= MOD:
        x -= MOD
    return x

class RollingHash:
    def __init__(self, max_len):
        # Random base in [1..MOD-2]
        seed = int(time.time_ns()) ^ random.getrandbits(64)
        rng = random.Random(seed)
        self.base = rng.randrange(1, MOD - 1)

        # Precompute powers of base up to max_len
        self.pow = [1] * (max_len + 1)
        for i in range(1, max_len + 1):
            self.pow[i] = mod_mul(self.pow[i - 1], self.base)

    def prefix_hash(self, s):
        # Build prefix hashes:
        # h[i+1] = hash(s[0..i]) (length i+1)
        h = [0] * (len(s) + 1)
        for i, ch in enumerate(s):
            h[i + 1] = mod_add(mod_mul(h[i], self.base), ord(ch))
        return h

    def range_hash(self, h, l, r):
        # hash of s[l:r] (half-open)
        # = h[r] - h[l] * base^(r-l)
        return mod_sub(h[r], mod_mul(h[l], self.pow[r - l]))

def solve():
    data = sys.stdin.read().strip().split()
    n, m = map(int, data[:2])
    A = data[2].strip()
    B = data[3].strip()

    # Double strings to model cyclic shifts as substrings
    A2 = A + A
    B2 = B + B

    # Prepare hashing (need powers up to 2*n, safe for all queries we do)
    rh = RollingHash(2 * n + 5)
    ha = rh.prefix_hash(A2)
    hb = rh.prefix_hash(B2)

    def lcp_a_b(i, j, max_len):
        """LCP between A2[i:] and B2[j:] up to max_len."""
        lo, hi = 0, max_len
        while lo < hi:
            mid = (lo + hi + 1) // 2
            if rh.range_hash(ha, i, i + mid) == rh.range_hash(hb, j, j + mid):
                lo = mid
            else:
                hi = mid - 1
        return lo

    def lcp_b_b(i, j, max_len):
        """LCP between B2[i:] and B2[j:] up to max_len."""
        lo, hi = 0, max_len
        while lo < hi:
            mid = (lo + hi + 1) // 2
            if rh.range_hash(hb, i, i + mid) == rh.range_hash(hb, j, j + mid):
                lo = mid
            else:
                hi = mid - 1
        return lo

    # Compare two cyclic shifts of B lexicographically
    def cmp_shift(i, j):
        l = lcp_b_b(i, j, m)
        if l == m:
            return False
        return B2[i + l] < B2[j + l]

    # Step 1: lexicographically smallest cyclic shift of B
    best_b = 0
    for i in range(1, m):
        if cmp_shift(i, best_b):
            best_b = i

    # Step 2: check if A2[pos:pos+m] differs from B2[best_b:best_b+m] by <= 1 char
    def mismatches_le_1(pos):
        l1 = lcp_a_b(pos, best_b, m)
        if l1 == m:
            return True  # 0 mismatches
        if l1 == m - 1:
            return True  # exactly 1 mismatch (last char)
        # skip one mismatch at l1, then remaining must match
        l2 = lcp_a_b(pos + l1 + 1, best_b + l1 + 1, m - l1 - 1)
        return (l1 + 1 + l2 == m)

    # Choose shift of A with minimal shift count; tie -> left shift.
    best_a = None
    for d in range(0, n // 2 + 1):
        # left shift by d => start index d
        if mismatches_le_1(d):
            best_a = d
            break
        # right shift by d => start index n-d (avoid duplicates)
        if d > 0 and n - d != d and mismatches_le_1(n - d):
            best_a = n - d
            break

    # Output results
    result_b = B2[best_b:best_b + m]
    result_a = A2[best_a:best_a + n]
    sys.stdout.write(result_b + "\n" + result_a + "\n")

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

---

## 5) Compressed editorial

- Double strings: `A2=A+A`, `B2=B+B`; cyclic shift = substring of fixed length.
- Use rolling hash to answer substring equality in O(1), and LCP via binary search in O(log L).
- Find minimal cyclic shift of `B` by scanning all starts `i∈[0..M-1]` and comparing `B2[i:i+M)` using `LCP` to locate first differing char.
- Let `B'` be that minimal shift.
- For each shift start `pos` in `A2`, decide if `A2[pos:pos+M)` matches `B'` with ≤1 mismatch:
  - `l1=LCP(start)`; if full match => ok; else skip one char and compute `l2`; if remainder matches => exactly 1 mismatch else ≥2.
- To satisfy tie-breaking on shifts of `A`, check shift distances `d=0..⌊N/2⌋`:
  - test left shift `d` first (`pos=d`), then right shift `d` (`pos=N-d`).
- Output `B2[best_b:best_b+M]` and `A2[best_a:best_a+N]`.