#include <bits/stdc++.h>
using namespace std;
short dp[505][505][505], nxt[505][2];
string a, b;
int main(int argc, char** argv) {
  int n, m;
  cin >> n >> m;
  cin >> a >> b;
  a = ' ' + a, b = ' ' + b;
  for (int i = 0; i <= m; i++) {
    string ling = b.substr(1, i) + '0';
    nxt[i][0] = 0;
    for (int j = min(m, (int)ling.size()); j >= 1; j--) {
      if (b.substr(1, j) == ling.substr(ling.size() - j)) {
        nxt[i][0] = j;
        break;
      }
    }
    string yi = b.substr(1, i) + '1';
    nxt[i][1] = 0;
    for (int j = min(m, (int)yi.size()); j >= 1; j--) {
      if (b.substr(1, j) == yi.substr(yi.size() - j)) {
        nxt[i][1] = j;
        break;
      }
    }
  }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++)
      for (int k = 0; k <= m; k++) dp[i][j][k] = 20000;
  dp[0][0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k <= m; k++) {
        dp[i + 1][j + (nxt[k][0] == m)][nxt[k][0]] =
            min(dp[i + 1][j + (nxt[k][0] == m)][nxt[k][0]],
                (short)(dp[i][j][k] + (a[i + 1] != '0')));
        dp[i + 1][j + (nxt[k][1] == m)][nxt[k][1]] =
            min(dp[i + 1][j + (nxt[k][1] == m)][nxt[k][1]],
                (short)(dp[i][j][k] + (a[i + 1] != '1')));
      }
    }
  }
  for (int i = 0; i <= n - m + 1; i++) {
    short ans = 20000;
    for (int k = 0; k <= m; k++) ans = min(ans, dp[n][i][k]);
    if (ans == 20000) ans = -1;
    cout << ans << " ";
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 namespace std;
short dp[505][505][505], nxt[505][2];
string a, b;
int main(int argc, char** argv) {
  int n, m;
  cin >> n >> m;
  cin >> a >> b;
  a = ' ' + a, b = ' ' + b;
  for (int i = 0; i <= m; i++) {
    string ling = b.substr(1, i) + '0';
    nxt[i][0] = 0;
    for (int j = min(m, (int)ling.size()); j >= 1; j--) {
      if (b.substr(1, j) == ling.substr(ling.size() - j)) {
        nxt[i][0] = j;
        break;
      }
    }
    string yi = b.substr(1, i) + '1';
    nxt[i][1] = 0;
    for (int j = min(m, (int)yi.size()); j >= 1; j--) {
      if (b.substr(1, j) == yi.substr(yi.size() - j)) {
        nxt[i][1] = j;
        break;
      }
    }
  }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++)
      for (int k = 0; k <= m; k++) dp[i][j][k] = 20000;
  dp[0][0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k <= m; k++) {
        dp[i + 1][j + (nxt[k][0] == m)][nxt[k][0]] =
            min(dp[i + 1][j + (nxt[k][0] == m)][nxt[k][0]],
                (short)(dp[i][j][k] + (a[i + 1] != '0')));
        dp[i + 1][j + (nxt[k][1] == m)][nxt[k][1]] =
            min(dp[i + 1][j + (nxt[k][1] == m)][nxt[k][1]],
                (short)(dp[i][j][k] + (a[i + 1] != '1')));
      }
    }
  }
  for (int i = 0; i <= n - m + 1; i++) {
    short ans = 20000;
    for (int k = 0; k <= m; k++) ans = min(ans, dp[n][i][k]);
    if (ans == 20000) ans = -1;
    cout << ans << " ";
  }
  return 0;
}

```