#include <bits/stdc++.h>
using namespace std;
const int INF = 1000;
int main() {
  int n, m;
  cin >> n >> m;
  string a;
  cin >> a;
  string b;
  cin >> b;
  vector<vector<int>> nxt(m + 1, vector<int>(2));
  for (int i = 0; i <= m; i++) {
    string p = b.substr(0, i);
    for (int j = 0; j < 2; j++) {
      string p2 = p + (char)('0' + j);
      int cnt = p2.size();
      for (int k = m; k >= 0; k--) {
        if (k <= cnt) {
          if (p2.substr(cnt - k) == b.substr(0, k)) {
            nxt[i][j] = k;
            break;
          }
        }
      }
    }
  }
  vector<vector<int>> dp(m + 1, vector<int>(n - m + 2, INF));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    vector<vector<int>> dp2(m + 1, vector<int>(n - m + 2, INF));
    for (int j = 0; j <= m; j++) {
      for (int k = 0; k <= n - m; k++) {
        if (dp[j][k] != INF) {
          if (a[i] == '0') {
            if (nxt[j][0] == m) {
              dp2[nxt[j][0]][k + 1] = min(dp2[nxt[j][0]][k + 1], dp[j][k]);
            } else {
              dp2[nxt[j][0]][k] = min(dp2[nxt[j][0]][k], dp[j][k]);
            }
            if (nxt[j][1] == m) {
              dp2[nxt[j][1]][k + 1] = min(dp2[nxt[j][1]][k + 1], dp[j][k] + 1);
            } else {
              dp2[nxt[j][1]][k] = min(dp2[nxt[j][1]][k], dp[j][k] + 1);
            }
          }
          if (a[i] == '1') {
            if (nxt[j][0] == m) {
              dp2[nxt[j][0]][k + 1] = min(dp2[nxt[j][0]][k + 1], dp[j][k] + 1);
            } else {
              dp2[nxt[j][0]][k] = min(dp2[nxt[j][0]][k], dp[j][k] + 1);
            }
            if (nxt[j][1] == m) {
              dp2[nxt[j][1]][k + 1] = min(dp2[nxt[j][1]][k + 1], dp[j][k]);
            } else {
              dp2[nxt[j][1]][k] = min(dp2[nxt[j][1]][k], dp[j][k]);
            }
          }
        }
      }
    }
    swap(dp, dp2);
  }
  for (int i = 0; i < n - m + 2; i++) {
    int ans = INF;
    for (int j = 0; j <= m; j++) {
      ans = min(ans, dp[j][i]);
    }
    if (ans == INF) {
      cout << -1;
    } else {
      cout << ans;
    }
    if (i < n - m + 1) {
      cout << ' ';
    }
  }
  cout << endl;
}

### Prompt

Your task is to create 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 namespace std;
const int INF = 1000;
int main() {
  int n, m;
  cin >> n >> m;
  string a;
  cin >> a;
  string b;
  cin >> b;
  vector<vector<int>> nxt(m + 1, vector<int>(2));
  for (int i = 0; i <= m; i++) {
    string p = b.substr(0, i);
    for (int j = 0; j < 2; j++) {
      string p2 = p + (char)('0' + j);
      int cnt = p2.size();
      for (int k = m; k >= 0; k--) {
        if (k <= cnt) {
          if (p2.substr(cnt - k) == b.substr(0, k)) {
            nxt[i][j] = k;
            break;
          }
        }
      }
    }
  }
  vector<vector<int>> dp(m + 1, vector<int>(n - m + 2, INF));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    vector<vector<int>> dp2(m + 1, vector<int>(n - m + 2, INF));
    for (int j = 0; j <= m; j++) {
      for (int k = 0; k <= n - m; k++) {
        if (dp[j][k] != INF) {
          if (a[i] == '0') {
            if (nxt[j][0] == m) {
              dp2[nxt[j][0]][k + 1] = min(dp2[nxt[j][0]][k + 1], dp[j][k]);
            } else {
              dp2[nxt[j][0]][k] = min(dp2[nxt[j][0]][k], dp[j][k]);
            }
            if (nxt[j][1] == m) {
              dp2[nxt[j][1]][k + 1] = min(dp2[nxt[j][1]][k + 1], dp[j][k] + 1);
            } else {
              dp2[nxt[j][1]][k] = min(dp2[nxt[j][1]][k], dp[j][k] + 1);
            }
          }
          if (a[i] == '1') {
            if (nxt[j][0] == m) {
              dp2[nxt[j][0]][k + 1] = min(dp2[nxt[j][0]][k + 1], dp[j][k] + 1);
            } else {
              dp2[nxt[j][0]][k] = min(dp2[nxt[j][0]][k], dp[j][k] + 1);
            }
            if (nxt[j][1] == m) {
              dp2[nxt[j][1]][k + 1] = min(dp2[nxt[j][1]][k + 1], dp[j][k]);
            } else {
              dp2[nxt[j][1]][k] = min(dp2[nxt[j][1]][k], dp[j][k]);
            }
          }
        }
      }
    }
    swap(dp, dp2);
  }
  for (int i = 0; i < n - m + 2; i++) {
    int ans = INF;
    for (int j = 0; j <= m; j++) {
      ans = min(ans, dp[j][i]);
    }
    if (ans == INF) {
      cout << -1;
    } else {
      cout << ans;
    }
    if (i < n - m + 1) {
      cout << ' ';
    }
  }
  cout << endl;
}

```