#include <bits/stdc++.h>
using ll = long long;
int const nmax = 500;
char v[5 + nmax], v2[5 + nmax];
int pre[5 + nmax], edges[5 + nmax][2];
void precompute(int m) {
  int ptr = 0;
  for (int i = 2; i <= m; i++) {
    while (true) {
      if (v2[i] == v2[1 + ptr]) {
        ptr++;
        pre[i] = ptr;
        break;
      } else if (0 == ptr) {
        pre[i] = 0;
        break;
      } else {
        ptr = pre[ptr];
      }
    }
  }
  for (int i = 0; i <= m; i++) {
    for (char ch = '0'; ch <= '1'; ch++) {
      int curr = i;
      while (true) {
        if (curr < m && v2[curr + 1] == ch) {
          curr++;
          break;
        } else if (0 < curr) {
          curr = pre[curr];
        } else {
          break;
        }
      }
      edges[i][ch - '0'] = curr;
    }
  }
}
int dp[5 + nmax][5 + nmax];
int dp2[5 + nmax][5 + nmax];
int main() {
  int n, m;
  std::cin >> n >> m;
  for (int i = 1; i <= n; i++) std::cin >> v[i];
  for (int i = 1; i <= m; i++) std::cin >> v2[i];
  precompute(m);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) dp[i][j] = dp2[i][j] = n * 2;
  dp[0][0] = 0;
  for (int step = 1; step <= n; step++) {
    int lim = std::min(step, m);
    int lim2 = std::max(0, step - m + 1);
    for (int i = 0; i <= lim2; i++)
      for (int j = 0; j <= lim; j++) {
        for (int h = 0; h < 2; h++) {
          dp2[i + (edges[j][h] == m)][edges[j][h]] =
              std::min(dp2[i + (edges[j][h] == m)][edges[j][h]],
                       dp[i][j] + (h != (v[step] - '0')));
        }
      }
    for (int i = 0; i <= lim2; i++)
      for (int j = 0; j <= lim; j++) {
        dp[i][j] = dp2[i][j];
        dp2[i][j] = n * 2;
      }
  }
  for (int i = 0; i <= n - m + 1; i++) {
    int result = dp[i][0];
    for (int j = 0; j <= m; j++) {
      result = std::min(result, dp[i][j]);
    }
    if (n < result) {
      std::cout << -1 << " ";
    } else
      std::cout << result << " ";
  }
  std::cout << '\n';
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

The Winter holiday will be here soon. Mr. Chanek wants to decorate his house's wall with ornaments. The wall can be represented as a binary string a of length n. His favorite nephew has another binary string b of length m (m ≤ n).

Mr. Chanek's nephew loves the non-negative integer k. His nephew wants exactly k occurrences of b as substrings in a. 

However, Mr. Chanek does not know the value of k. So, for each k (0 ≤ k ≤ n - m + 1), find the minimum number of elements in a that have to be changed such that there are exactly k occurrences of b in a.

A string s occurs exactly k times in t if there are exactly k different pairs (p,q) such that we can obtain s by deleting p characters from the beginning and q characters from the end of t.

Input

The first line contains two integers n and m (1 ≤ m ≤ n ≤ 500) — size of the binary string a and b respectively.

The second line contains a binary string a of length n.

The third line contains a binary string b of length m.

Output

Output n - m + 2 integers — the (k+1)-th integer denotes the minimal number of elements in a that have to be changed so there are exactly k occurrences of b as a substring in a.

Example

Input


9 3
100101011
101


Output


1 1 0 1 6 -1 -1 -1

Note

For k = 0, to make the string a have no occurrence of 101, you can do one character change as follows.

100101011 → 100100011

For k = 1, you can also change a single character.

100101011 → 100001011

For k = 2, no changes are needed.

### Response

```cpp
#include <bits/stdc++.h>
using ll = long long;
int const nmax = 500;
char v[5 + nmax], v2[5 + nmax];
int pre[5 + nmax], edges[5 + nmax][2];
void precompute(int m) {
  int ptr = 0;
  for (int i = 2; i <= m; i++) {
    while (true) {
      if (v2[i] == v2[1 + ptr]) {
        ptr++;
        pre[i] = ptr;
        break;
      } else if (0 == ptr) {
        pre[i] = 0;
        break;
      } else {
        ptr = pre[ptr];
      }
    }
  }
  for (int i = 0; i <= m; i++) {
    for (char ch = '0'; ch <= '1'; ch++) {
      int curr = i;
      while (true) {
        if (curr < m && v2[curr + 1] == ch) {
          curr++;
          break;
        } else if (0 < curr) {
          curr = pre[curr];
        } else {
          break;
        }
      }
      edges[i][ch - '0'] = curr;
    }
  }
}
int dp[5 + nmax][5 + nmax];
int dp2[5 + nmax][5 + nmax];
int main() {
  int n, m;
  std::cin >> n >> m;
  for (int i = 1; i <= n; i++) std::cin >> v[i];
  for (int i = 1; i <= m; i++) std::cin >> v2[i];
  precompute(m);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) dp[i][j] = dp2[i][j] = n * 2;
  dp[0][0] = 0;
  for (int step = 1; step <= n; step++) {
    int lim = std::min(step, m);
    int lim2 = std::max(0, step - m + 1);
    for (int i = 0; i <= lim2; i++)
      for (int j = 0; j <= lim; j++) {
        for (int h = 0; h < 2; h++) {
          dp2[i + (edges[j][h] == m)][edges[j][h]] =
              std::min(dp2[i + (edges[j][h] == m)][edges[j][h]],
                       dp[i][j] + (h != (v[step] - '0')));
        }
      }
    for (int i = 0; i <= lim2; i++)
      for (int j = 0; j <= lim; j++) {
        dp[i][j] = dp2[i][j];
        dp2[i][j] = n * 2;
      }
  }
  for (int i = 0; i <= n - m + 1; i++) {
    int result = dp[i][0];
    for (int j = 0; j <= m; j++) {
      result = std::min(result, dp[i][j]);
    }
    if (n < result) {
      std::cout << -1 << " ";
    } else
      std::cout << result << " ";
  }
  std::cout << '\n';
  return 0;
}

```