## 1) Abridged problem statement

You have a binary row of **N** coins (0=heads, 1=tails). There is an unknown binary vector **A** of length **K−1** that defines a transformation **X** on any length-**K** block:

1. Cyclically shift the block left by 1.
2. For each position `i = 1..K−1` (left to right): if the coin at position `i` is `1` and `A[i]=1`, then flip the **last** coin of the block.

You are given **L** example pairs (before, after) for transformation **X**, and it’s guaranteed there is **exactly one** vector **A** consistent with all examples.

Then **M** operations were applied to an unknown initial length-**N** row: each operation chooses a consecutive length-**K** subrow starting at `S_i` and applies transformation **X** exactly `D_i` times to that subrow.

You are given the final row after all operations; reconstruct and output the **initial** row.

Constraints: `N<=50, M<=10, L<=200, K<=N, D_i<=1e6`.

---

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

### A. Understand transformation X algebraically (over GF(2))

Let the current length-**K** block be bits `C[0..K-1]` (0-based, left→right). Vector `A` has bits `A[0..K-2]`.

**Step 1: cyclic left shift**
After shift, bit positions become:
- `S[j] = C[(j+1) mod K]`

So:
- `S[0] = C[1]`, …, `S[K-2]=C[K-1]`, `S[K-1]=C[0]`.

**Step 2: conditional flips of last coin**
The last bit `S[K-1]` is flipped once for each `i=0..K-2` where `S[i]=1` and `A[i]=1`.

Over GF(2), “flip” is XOR with 1, and repeated flips XOR together:
\[
\text{newLast} = S[K-1] \oplus \bigoplus_{i=0}^{K-2} (S[i] \land A[i])
\]
All other positions `0..K-2` remain `S[0..K-2]`.

So the transformation is linear in the unknowns `A[i]`, once `before/after` are fixed.

---

### B. Reconstruct A using the L constraints (Gaussian elimination in GF(2))

Each constraint gives:
- `before` block `B` (K bits)
- `after` block `T` (K bits)

From the transformation:
- The first `K-1` bits of `after` equal `before` shifted left: indeed `after[i] = B[i+1]` for `i=0..K-2`. (These serve as consistency checks but not needed.)
- The only equation involving `A` comes from the last bit:

Let:
- `S[i]=B[i+1]` for `i=0..K-2`
- `S[K-1]=B[0]`

Then:
\[
T[K-1] = B[0] \oplus \bigoplus_{i=0}^{K-2} (B[i+1] \land A[i])
\]

Rearrange:
\[
\bigoplus_{i=0}^{K-2} (B[i+1] \land A[i]) = T[K-1] \oplus B[0]
\]

That is a linear equation in unknown bits `A[0..K-2]`.
For each constraint we build:
- coefficient vector `coeff[i] = B[i+1]`
- RHS `rhs = T[K-1] XOR B[0]`

Now we have up to `L<=200` equations in `K-1 <= 49` unknowns. Use Gaussian elimination over GF(2) with 64-bit masks.

Representation:
- store coefficients as a `uint64_t` mask (bit `i` is coeff of `A[i]`)
- store RHS as an `int` (0/1)

Perform elimination to reduced row-echelon style:
- for each column, find a row with that bit set, pivot it, XOR-eliminate from all other rows.
Because the problem guarantees **unique** solution, the resulting system yields a single `A`.

Extract solution:
- after elimination, each nonzero row has a leading 1 at some column `c` and the row means `A[c] = rhs` (because all other bits in that row have been eliminated away).

---

### C. Undo all operations by applying the inverse of X

We are given the final length-**N** row. Operations were applied forward in order 1..M, each applying X `D_i` times to a block.

To recover the initial row, we must reverse operations in order **M..1** and apply **X^{-1}** `D_i` times.

#### Key trick: X is easy to invert
Let’s denote:
- `after` block = `Y`
- `before` block = `B`

From Step 1 (left shift), we know:
- `Y[0..K-2] = B[1..K-1]`
So if we already have `Y`, then we already know `B[1..K-1]` directly.

The only unknown is `B[0]`.

From last-bit relation:
\[
Y[K-1] = B[0] \oplus \bigoplus_{i=0}^{K-2} (B[i+1] \land A[i])
\]
But `B[i+1] = Y[i]` for `i=0..K-2`, so:
\[
B[0] = Y[K-1] \oplus \bigoplus_{i=0}^{K-2} (Y[i] \land A[i])
\]

So given `Y` and `A`, we can compute `B[0]` and thus reconstruct the whole previous block:
- `B[1..K-1] = Y[0..K-2]`
- `B[0]` computed by parity.

This is exactly what the C++ code does: for each iteration it takes the current block (`sub`), computes the needed flip parity from its lower `K-1` bits, fixes the top bit, then rotates right by 1 to undo the previous left shift.

#### Complexity
`N<=50`, `M<=10`, but `D_i` can be large (1e6). The provided solution still loops `sum(D_i)` times, but with tiny constant cost: bit extraction, popcount, shifts on 64-bit integers. With the contest limits, this is intended to pass.

---

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

```cpp
#include <bits/stdc++.h>              // Includes almost all standard headers (GNU extension)

using namespace std;

// Helper: print a pair as "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

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

// Helper: read a whole vector (assumes correct size already)
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Helper: print a vector (space-separated)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, m, k, l;                                   // N coins, M operations, block size K, L constraints
vector<pair<int, int>> ops;                       // operations: (start index s, repetitions d)
vector<pair<uint64_t, uint64_t>> constraints;     // each constraint: (beforeMask, afterMask) for K bits
uint64_t final_row;                               // final N-bit row as a bitmask (bit i = coin i)

void read() {
    cin >> n >> m >> k >> l;

    ops.resize(m);
    for(auto& [s, d]: ops) {
        cin >> s >> d;    // input is 1-based S
        s--;              // convert to 0-based index
    }

    constraints.resize(l);
    for(auto& [before, after]: constraints) {
        string sb, sa;
        cin >> sb >> sa;  // K-bit strings: before and after

        before = 0;
        after = 0;

        // Convert each string into a bitmask: bit i corresponds to position i in the block
        for(int i = 0; i < k; i++) {
            if(sb[i] == '1') {
                before |= (1ull << i);
            }
            if(sa[i] == '1') {
                after |= (1ull << i);
            }
        }
    }

    // Read final N-bit row and store as bitmask
    string s;
    cin >> s;
    final_row = 0;
    for(int i = 0; i < n; i++) {
        if(s[i] == '1') {
            final_row |= (1ull << i);
        }
    }
}

void solve() {
    // Build linear equations for unknown vector A (length K-1).
    // For each constraint (before, after):
    //   XOR_{i=0..K-2} (before[i+1] * A[i]) = after[K-1] XOR before[0]
    // Store coefficients as a uint64 mask and RHS as a bit.
    vector<pair<uint64_t, int>> eqs(l);
    for(int i = 0; i < l; i++) {
        auto [before, after] = constraints[i];

        // coefficients for A are bits before[1..K-1], shifted down to positions 0..K-2
        uint64_t coeffs = (before >> 1);

        // rhs = after[last] XOR before[first]
        // after[last] is bit (k-1)
        // before[first] is bit 0
        int rhs = ((after >> (k - 1)) ^ (before & 1)) & 1;

        eqs[i] = {coeffs, rhs};
    }

    // Gaussian elimination over GF(2) on variables A[0..K-2]
    uint64_t a = 0;         // will store recovered A bits
    int pivot_row = 0;      // current row to place pivot

    for(int col = 0; col < k - 1 && pivot_row < l; col++) {
        int found = -1;

        // Find a row with a 1 in this column at or below pivot_row
        for(int row = pivot_row; row < l; row++) {
            if(eqs[row].first & (1ull << col)) {
                found = row;
                break;
            }
        }
        if(found == -1) {
            continue;       // no pivot in this column
        }

        // Move found pivot row into position pivot_row
        swap(eqs[pivot_row], eqs[found]);

        // Eliminate this column from all other rows (full elimination -> RREF-like)
        for(int row = 0; row < l; row++) {
            if(row != pivot_row && (eqs[row].first & (1ull << col))) {
                eqs[row].first ^= eqs[pivot_row].first;   // XOR coefficients
                eqs[row].second ^= eqs[pivot_row].second; // XOR RHS
            }
        }
        pivot_row++;
    }

    // Extract solution: each non-zero row should represent one variable = rhs
    for(int i = 0; i < l; i++) {
        if(eqs[i].first == 0) {
            continue; // either redundant or inconsistent (but problem guarantees solvable)
        }

        // Find index of least significant set bit (pivot column)
        int col = __builtin_ctzll(eqs[i].first);

        // In reduced form, that variable equals rhs
        if(eqs[i].second) {
            a |= (1ull << col);
        }
    }

    // Now invert all operations starting from final_row
    uint64_t mask = (1ull << k) - 1;            // mask for K bits
    uint64_t row = final_row;                   // current reconstructed row (we will go backwards)
    uint64_t lower_bits = (1ull << (k - 1)) - 1;// mask for lowest K-1 bits within a K-bit block

    // Reverse operations: last to first
    for(int op = m - 1; op >= 0; op--) {
        int s = ops[op].first;   // start index of block in the N-bit row
        int d = ops[op].second;  // number of times X was applied forward

        // To undo, apply X^{-1} d times
        for(int t = 0; t < d; t++) {
            // Extract the current K-bit block (this is "after" block for inverse step)
            uint64_t sub = (row >> s) & mask;

            // Compute parity of (sub[i] & a[i]) over i=0..K-2
            // Only lower K-1 bits of sub participate; top bit is separate.
            int flip_bit = __builtin_popcountll((sub & a) & lower_bits) & 1;

            // We want to reconstruct the shifted block S (after left shift, after flips).
            // In inverse, we compute original B[0] by:
            //   B[0] = sub[K-1] XOR parity(sub[0..K-2] & a)
            // This code effectively "fixes" the top bit accordingly:
            uint64_t shifted = sub ^ ((uint64_t)flip_bit << (k - 1));

            // Undo the cyclic left shift: original B is shifted rotated right by 1:
            // B = rotr(shifted, 1) within K bits
            uint64_t original = ((shifted << 1) | (shifted >> (k - 1))) & mask;

            // Write the restored block back into the N-bit row
            row = (row & ~(mask << s)) | (original << s);
        }
    }

    // Output reconstructed initial row as N bits (bit i -> character i)
    for(int i = 0; i < n; i++) {
        cout << ((row >> i) & 1);
    }
    cout << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);  // fast I/O
    cin.tie(nullptr);

    int T = 1;                         // single test case (no multiple tests in input)
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach, heavily commented)

```python
import sys

def read_ints():
    return list(map(int, sys.stdin.readline().split()))

def solve():
    # Read N, M, K, L
    n, m, k, l = read_ints()

    # Read M operations (Si, Di); convert Si to 0-based
    ops_raw = read_ints()
    ops = []
    for i in range(m):
        s = ops_raw[2*i] - 1
        d = ops_raw[2*i + 1]
        ops.append((s, d))

    # Read L constraints: each is two K-bit strings (before, after)
    constraints = []
    for _ in range(l):
        sb = sys.stdin.readline().strip()
        sa = sys.stdin.readline().strip()

        before = 0
        after = 0
        # Convert strings into bitmasks, bit i corresponds to position i
        for i, ch in enumerate(sb):
            if ch == '1':
                before |= 1 << i
        for i, ch in enumerate(sa):
            if ch == '1':
                after |= 1 << i

        constraints.append((before, after))

    # Read final N-bit row
    s_final = sys.stdin.readline().strip()
    final_row = 0
    for i, ch in enumerate(s_final):
        if ch == '1':
            final_row |= 1 << i

    # -------------------------
    # 1) Reconstruct vector A (K-1 unknown bits) using Gaussian elimination over GF(2)
    # Each equation:
    #   XOR_{i=0..K-2} (before[i+1] & A[i]) = after[K-1] XOR before[0]
    # Coeff mask is before>>1 (bits 0..K-2)
    # -------------------------
    eqs = []
    for before, after in constraints:
        coeff = before >> 1
        rhs = ((after >> (k - 1)) & 1) ^ (before & 1)
        eqs.append([coeff, rhs])

    # Gaussian elimination in GF(2) for columns 0..K-2
    pivot_row = 0
    for col in range(k - 1):
        if pivot_row >= l:
            break

        # Find pivot row with this column bit set
        found = -1
        for r in range(pivot_row, l):
            if (eqs[r][0] >> col) & 1:
                found = r
                break
        if found == -1:
            continue

        # Swap into pivot position
        eqs[pivot_row], eqs[found] = eqs[found], eqs[pivot_row]

        # Eliminate this column from all other rows
        piv_coeff, piv_rhs = eqs[pivot_row]
        for r in range(l):
            if r != pivot_row and ((eqs[r][0] >> col) & 1):
                eqs[r][0] ^= piv_coeff
                eqs[r][1] ^= piv_rhs

        pivot_row += 1

    # Extract A from reduced system
    a = 0
    for coeff, rhs in eqs:
        if coeff == 0:
            continue
        # index of least significant set bit
        col = (coeff & -coeff).bit_length() - 1
        if rhs:
            a |= 1 << col

    # -------------------------
    # 2) Reverse operations by applying X^{-1} repeatedly to each affected K-block
    # -------------------------
    mask_k = (1 << k) - 1           # mask to keep K bits
    lower_mask = (1 << (k - 1)) - 1 # mask for bits 0..K-2 inside a K-bit block

    row = final_row

    # Process operations in reverse order
    for s, d in reversed(ops):
        for _ in range(d):
            # sub is the current K-bit block (this corresponds to "after" in inverse)
            sub = (row >> s) & mask_k

            # parity of (sub[0..K-2] & a)
            # Python: bit_count() is popcount
            flip = (((sub & a) & lower_mask).bit_count() & 1)

            # Fix top bit to recover the shifted state before undoing rotation
            shifted = sub ^ (flip << (k - 1))

            # Undo the cyclic left shift (rotate right by 1 within K bits)
            original = ((shifted << 1) | (shifted >> (k - 1))) & mask_k

            # Put block back into row
            row = (row & ~(mask_k << s)) | (original << s)

    # Output row as N bits (bit i corresponds to position i)
    out = []
    for i in range(n):
        out.append('1' if ((row >> i) & 1) else '0')
    sys.stdout.write(''.join(out) + "\n")

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

---

## 5) Compressed editorial

- Encode each length-**K** string as a bitmask.
- For each constraint `(before, after)`, only the last bit depends on unknown **A**:
  \[
  \bigoplus_{i=0}^{K-2} (before[i+1]\land A[i]) = after[K-1] \oplus before[0]
  \]
  Thus build `L` linear equations in `K−1` unknown bits; solve uniquely by Gaussian elimination over GF(2) using 64-bit masks.
- To reconstruct the initial row, reverse the operations. Inverse of **X** on a K-block is easy:
  - Given `after = Y`, we know `before[1..K-1] = Y[0..K-2]`.
  - Compute `before[0] = Y[K-1] XOR parity(Y[0..K-2] & A)`.
  - This corresponds to fixing the top bit then rotating right by 1.
- Apply this inverse step `D_i` times for each operation in reverse order, update the N-bit row bitmask, finally print the recovered initial bits.