<|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.

173. Coins
time limit per test: 0.75 sec.
memory limit per test: 4096 KB
input: standard
output: standard



There is a row consist of N coins. Each coin lies heads up or tails up. There is also binary vector A with (K-1) dimensions.
We accomplish M operations with our coins row: we select some consecutive K coins, and apply to selected coins transformation X for Di times (i=1..M, where i is a number of operation)
The transformation X for given row C with length K means:
1. We shift the row C in one position to the left in a cyclic way.
2. Then we scan the row C from the first (leftmost) coin to (K-1)-th coin: if coin Ci lies tails up, and Ai=1, then we turn over coin Ck
Obviously, that for constant coins row with length K transformation X is fully determined by vector A. But vector A haven't given to you. You have results of transformation L rows with length K (row of coins before transformation, and row after transformation). It is guarantied that there is exactly one vector A satisfied all L conditions.
Your task is to reconstruct initial row by given row after all operations being accomplished.

Input
There is four integer numbers in the first line of input: N, M, K, L (2<=N<=50, 1<=M<=10, 1<=L<=200, 2<=K<=N).
Second line contains description of accomplished operations, M pairs of positive integer numbers Si and Di. Si is a starting point of selected coins subrow, Di is a number of iterations of transformation (Si <= N-K+1; Di <= 10^6).
There are the following L conditions determining transformation X, L blocks by two lines each: first line of each block contains a row before transformation, and the second line contains row after transformation. Each line consists of K symbols, 1 denotes tails, 0 denotes heads.
The last line contains N symbols - row of coins after all operations accomplished. i-th symbol is 1 if i-th coin lies tails up, and 0 otherwise.

Output
Write N symbols in single line of output: i-th symbol must be 1 if i-th coin of initial row laid tails up, and 0 otherwise.

Sample test(s)

Input
4 2 3 2
2 1 1 1
010
101
101
011
1010

Output
0110
Author:	Dmitry Orlov
Resource:	Saratov ST team Spring Contest #1
Date:	18.05.2003

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

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

1. Cyclic 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.

You are given **L** examples of `(beforeBlock, afterBlock)` for **X**, and it is guaranteed there is **exactly one** vector **A** consistent with all examples.

Then **M** operations were applied to an unknown initial length-**N** row: operation `i` takes the block starting at `S_i` (length K) and applies **X** exactly `D_i` times. You are given the final row; output the initial row.

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

---

## 2) Key observations

1. **Only the last bit depends on A** (for a single application of X).  
   After the cyclic shift, positions `0..K−2` are determined purely by the shift. The conditional flips only affect the last coin.

2. **Each example yields one linear equation over GF(2)** (XOR arithmetic).  
   For an example `(before=B, after=T)`, the last bit gives:
   \[
   \bigoplus_{i=0}^{K-2} (B[i+1]\land A[i]) = T[K-1]\oplus B[0]
   \]
   This is a linear equation in the unknown bits of `A`.

3. Since `K−1 <= 49`, we can represent equations as **64-bit masks** and solve with **Gaussian elimination over GF(2)**.

4. To reconstruct the initial row, we must **reverse operations** and apply **X^{-1}** repeatedly.  
   Inversion is easy:
   - From `after` block `Y`, we know `before[1..K-1] = Y[0..K-2]`.
   - Compute `before[0]` using the last-bit relation:
     \[
     before[0] = Y[K-1] \oplus \bigoplus_{i=0}^{K-2}(Y[i]\land A[i])
     \]
   Then we “rotate right by 1” to undo the previous left rotation.

---

## 3) Full solution approach

### Step A: Parse input and encode bitstrings as bitmasks
- Encode each length-`K` block as a `uint64_t` mask where bit `i` corresponds to position `i` in the block (left to right).
- Encode the final length-`N` row similarly (still fits in 64-bit since `N<=50`).

### Step B: Build a linear system to recover A
Unknowns: `A[0..K-2]` (total `K-1` bits).

For each constraint `(before=B, after=T)`:
- Coefficients for `A[i]` are `B[i+1]`. As a bitmask this is `B >> 1` (bit 0 corresponds to `B[1]`, etc.).
- Right-hand side: `rhs = T[K-1] XOR B[0]`.

So each equation is:
- `mask = (B >> 1)` over variables
- `rhs` is 0/1

Run Gaussian elimination over GF(2) on these `(mask, rhs)` pairs. The statement guarantees a **unique solution**, so we can extract `A` directly from the reduced system.

### Step C: Reverse the M operations using X^{-1}
We are given the final row. To get the initial row:
- Process operations in reverse order: `i = M..1`.
- For each operation, apply the inverse transform `D_i` times to the affected block.

One inverse iteration on a K-bit block `sub` (this is the “after” of forward X):
1. Compute `flipParity = parity( (sub[0..K-2] & A) )`.
2. `before0 = sub[K-1] XOR flipParity`.
3. Reconstruct the “shifted” state by setting last bit accordingly, then rotate right by 1 to undo the left rotation.

Implementation-wise (matching the reference code):
- Extract `sub = (row >> s) & maskK`.
- `flipParity = popcount((sub & A) & lowerMask) % 2`
- `shifted = sub XOR (flipParity << (K-1))`
- `original = rotR1(shifted)` within K bits
- Write back into the big row mask.

Finally, print the recovered mask as `N` characters.

Complexity:
- Gaussian elimination: `O(L*(K-1)^2)` with small constants.
- Simulation: `sum(D_i)` iterations, each O(1) bit ops; with `M<=10, D_i<=1e6`, worst-case ~1e7 iterations, intended to pass in C++ under given limits.

---

## 4) C++ implementation

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

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

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

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

int n, m, k, l;
vector<pair<int, int>> ops;
vector<pair<uint64_t, uint64_t>> constraints;
uint64_t final_row;

void read() {
    cin >> n >> m >> k >> l;
    ops.resize(m);
    for(auto& [s, d]: ops) {
        cin >> s >> d;
        s--;
    }
    constraints.resize(l);
    for(auto& [before, after]: constraints) {
        string sb, sa;
        cin >> sb >> sa;
        before = 0;
        after = 0;
        for(int i = 0; i < k; i++) {
            if(sb[i] == '1') {
                before |= (1ull << i);
            }
            if(sa[i] == '1') {
                after |= (1ull << i);
            }
        }
    }
    string s;
    cin >> s;
    final_row = 0;
    for(int i = 0; i < n; i++) {
        if(s[i] == '1') {
            final_row |= (1ull << i);
        }
    }
}

void solve() {
    // The problem statement is a bit convoluted, but the key is that there is a
    // single transformation X that is valid based on the L constraints.
    // Particularly, each of the L constraint can be formalized as a linear
    // equation in Z2 - whenever A and X[1:K] match, we flip X[0] (which will
    // later be placed at X[-1] due to the cyclic shift). The variables are
    // given by A, while we already have the coefficients.
    //
    // The constraint is that L <= 200, so we can do an initial Gauss
    // elimination to find A. We don't have to, but we can use bitsets / int64_t
    // for it too to make it quicker. Afterwards, we simply need to simulate the
    // operations.
    //
    // Simulating might be a bit slow if done naively as Di <= 10^6 and we have
    // M <= 10 of these, so we should try to make each individual update
    // quickly. This can be done quickly if we keep the initial sequence and the
    // A as a bitmask / int64_t too, as then the cyclic shift and X
    // transformation ends up being a few bit operations plus a popcount which
    // is quick.

    vector<pair<uint64_t, int>> eqs(l);
    for(int i = 0; i < l; i++) {
        auto [before, after] = constraints[i];
        uint64_t coeffs = (before >> 1);
        int rhs = ((after >> (k - 1)) ^ (before & 1)) & 1;
        eqs[i] = {coeffs, rhs};
    }

    uint64_t a = 0;
    int pivot_row = 0;
    for(int col = 0; col < k - 1 && pivot_row < l; col++) {
        int found = -1;
        for(int row = pivot_row; row < l; row++) {
            if(eqs[row].first & (1ull << col)) {
                found = row;
                break;
            }
        }
        if(found == -1) {
            continue;
        }
        swap(eqs[pivot_row], eqs[found]);
        for(int row = 0; row < l; row++) {
            if(row != pivot_row && (eqs[row].first & (1ull << col))) {
                eqs[row].first ^= eqs[pivot_row].first;
                eqs[row].second ^= eqs[pivot_row].second;
            }
        }
        pivot_row++;
    }

    for(int i = 0; i < l; i++) {
        if(eqs[i].first == 0) {
            continue;
        }
        int col = __builtin_ctzll(eqs[i].first);
        if(eqs[i].second) {
            a |= (1ull << col);
        }
    }

    uint64_t mask = (1ull << k) - 1;
    uint64_t row = final_row;
    uint64_t lower_bits = (1ull << (k - 1)) - 1;
    for(int op = m - 1; op >= 0; op--) {
        int s = ops[op].first;
        int d = ops[op].second;
        for(int t = 0; t < d; t++) {
            uint64_t sub = (row >> s) & mask;
            int flip_bit = __builtin_popcountll((sub & a) & lower_bits) & 1;
            uint64_t shifted = sub ^ ((uint64_t)flip_bit << (k - 1));
            uint64_t original = ((shifted << 1) | (shifted >> (k - 1))) & mask;
            row = (row & ~(mask << s)) | (original << s);
        }
    }

    for(int i = 0; i < n; i++) {
        cout << ((row >> i) & 1);
    }
    cout << '\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();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 5) Python implementation (detailed comments)

```python
import sys

# p173 - Coins (Python)
# Same approach:
# 1) Solve for A with Gaussian elimination over GF(2) using integers as bitmasks.
# 2) Undo operations in reverse by applying X^{-1} Di times to the K-bit block.

def solve():
    input = sys.stdin.readline
    N, M, K, L = map(int, input().split())

    # Read 2*M integers (S_i, D_i)
    ops_raw = list(map(int, input().split()))
    ops = []
    for i in range(M):
        s = ops_raw[2*i] - 1  # to 0-based
        d = ops_raw[2*i + 1]
        ops.append((s, d))

    # Read constraints
    constraints = []
    for _ in range(L):
        sb = input().strip()
        sa = input().strip()
        before = 0
        after = 0
        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
    sfinal = input().strip()
    row = 0
    for i, ch in enumerate(sfinal):
        if ch == '1':
            row |= 1 << i

    # -------------------------
    # Build equations for A[0..K-2]:
    # (before>>1) dot A = after[K-1] XOR before[0]
    # -------------------------
    eqs = []
    for before, after in constraints:
        coeff = before >> 1
        rhs = ((after >> (K - 1)) & 1) ^ (before & 1)
        eqs.append([coeff, rhs])

    # -------------------------
    # Gaussian elimination over GF(2)
    # We eliminate pivot column from all other rows (RREF-ish).
    # -------------------------
    pivot_row = 0
    for col in range(K - 1):
        if pivot_row >= L:
            break

        # find pivot
        sel = -1
        for r in range(pivot_row, L):
            if (eqs[r][0] >> col) & 1:
                sel = r
                break
        if sel == -1:
            continue

        eqs[pivot_row], eqs[sel] = eqs[sel], eqs[pivot_row]
        piv_mask, piv_rhs = eqs[pivot_row]

        # eliminate from all other rows
        for r in range(L):
            if r != pivot_row and ((eqs[r][0] >> col) & 1):
                eqs[r][0] ^= piv_mask
                eqs[r][1] ^= piv_rhs

        pivot_row += 1

    # Extract A (unique solution guaranteed)
    A = 0
    for mask, rhs in eqs:
        if mask == 0:
            continue
        lead = (mask & -mask).bit_length() - 1  # index of lsb set bit
        if rhs:
            A |= 1 << lead

    # -------------------------
    # Undo operations using inverse transform
    # -------------------------
    maskK = (1 << K) - 1
    lowerMask = (1 << (K - 1)) - 1

    for s, d in reversed(ops):
        for _ in range(d):
            sub = (row >> s) & maskK

            # parity of (sub[0..K-2] & A)
            flip_parity = (((sub & A) & lowerMask).bit_count() & 1)

            shifted = sub ^ (flip_parity << (K - 1))

            # rotate right by 1 within K bits (same bit layout as in C++ solution)
            original = ((shifted << 1) | (shifted >> (K - 1))) & maskK

            row = (row & ~(maskK << s)) | (original << s)

    # Output N bits
    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()
```

If you want, I can also show a small worked example of how one constraint becomes a GF(2) equation, or how the inverse step reconstructs `before[0]` from the “after” block.