#include <bits/stdc++.h>
const int N = 505;
using namespace std;
int n;
int m;
int a[N];
int b[N];
int p[N];
int f[N][2];
int d[N][N];
int d2[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    char c;
    cin >> c;
    a[i] = c - '0';
  }
  for (int i = 0; i < m; i++) {
    char c;
    cin >> c;
    b[i] = c - '0';
  }
  for (int i = 1; i < m; i++) {
    p[i] = p[i - 1];
    while (p[i] > 0 && b[p[i]] != b[i]) {
      p[i] = p[p[i] - 1];
    }
    if (b[p[i]] == b[i]) {
      p[i] += 1;
    }
  }
  for (int i = 0; i < m; i++) {
    for (int x = 0; x < 2; x++) {
      if (i > 0 && x != b[i]) {
        f[i][x] = f[p[i - 1]][x];
      } else {
        f[i][x] = i + (x == b[i]);
      }
    }
  }
  f[m][0] = f[p[m - 1]][0];
  f[m][1] = f[p[m - 1]][1];
  int inf = 1e9;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      d[i][j] = d2[i][j] = inf;
    }
  }
  d[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= m; j++) {
      for (int h = 0; h <= n; h++) {
        for (int x = 0; x < 2; x++) {
          d2[f[j][x]][(f[j][x] == m) + h] =
              min(d2[f[j][x]][(f[j][x] == m) + h], d[j][h] + (x ^ a[i]));
        }
      }
    }
    for (int j = 0; j < N; j++) {
      for (int h = 0; h < N; h++) {
        d[j][h] = d2[j][h];
        d2[j][h] = inf;
      }
    }
  }
  for (int i = 0; i <= n - m + 1; i++) {
    int res = inf;
    for (int j = 0; j < m + 1; j++) {
      res = min(res, d[j][i]);
    }
    if (res == inf) {
      res = -1;
    }
    cout << res << " ";
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
const int N = 505;
using namespace std;
int n;
int m;
int a[N];
int b[N];
int p[N];
int f[N][2];
int d[N][N];
int d2[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    char c;
    cin >> c;
    a[i] = c - '0';
  }
  for (int i = 0; i < m; i++) {
    char c;
    cin >> c;
    b[i] = c - '0';
  }
  for (int i = 1; i < m; i++) {
    p[i] = p[i - 1];
    while (p[i] > 0 && b[p[i]] != b[i]) {
      p[i] = p[p[i] - 1];
    }
    if (b[p[i]] == b[i]) {
      p[i] += 1;
    }
  }
  for (int i = 0; i < m; i++) {
    for (int x = 0; x < 2; x++) {
      if (i > 0 && x != b[i]) {
        f[i][x] = f[p[i - 1]][x];
      } else {
        f[i][x] = i + (x == b[i]);
      }
    }
  }
  f[m][0] = f[p[m - 1]][0];
  f[m][1] = f[p[m - 1]][1];
  int inf = 1e9;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      d[i][j] = d2[i][j] = inf;
    }
  }
  d[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= m; j++) {
      for (int h = 0; h <= n; h++) {
        for (int x = 0; x < 2; x++) {
          d2[f[j][x]][(f[j][x] == m) + h] =
              min(d2[f[j][x]][(f[j][x] == m) + h], d[j][h] + (x ^ a[i]));
        }
      }
    }
    for (int j = 0; j < N; j++) {
      for (int h = 0; h < N; h++) {
        d[j][h] = d2[j][h];
        d2[j][h] = inf;
      }
    }
  }
  for (int i = 0; i <= n - m + 1; i++) {
    int res = inf;
    for (int j = 0; j < m + 1; j++) {
      res = min(res, d[j][i]);
    }
    if (res == inf) {
      res = -1;
    }
    cout << res << " ";
  }
}

```