#include <bits/stdc++.h>
using namespace std;
const long long inf = 2e9;
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  long long n, m;
  string s, t;
  cin >> n >> m >> s >> t;
  vector<long long> pi(m, 0), nxt(m, 0);
  for (long long i = 1; i < m; i++) {
    {
      t[i] = char('0' + (t[i] - '0') ^ 1);
      long long j = pi[i - 1];
      while (j > 0 and t[i] != t[j]) {
        j = pi[j - 1];
      }
      if (t[i] == t[j]) j++;
      nxt[i] = j;
      t[i] = char('0' + (t[i] - '0') ^ 1);
    }
    {
      long long j = pi[i - 1];
      while (j > 0 and t[i] != t[j]) {
        j = pi[j - 1];
      }
      if (t[i] == t[j]) j++;
      pi[i] = j;
    }
  }
  s = '#' + s;
  long long dp[n + 1][m][n - m + 2];
  fill(**dp, **dp + sizeof dp / sizeof(long long), inf);
  dp[0][0][0] = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      for (long long k = 0; k < n - m + 2; k++) {
        if (dp[i][j][k] == inf) continue;
        if (j == m - 1) {
          dp[i + 1][pi.back()][k + 1] = min(dp[i + 1][pi.back()][k + 1],
                                            dp[i][j][k] + (s[i + 1] != t[j]));
        } else {
          dp[i + 1][j + 1][k] =
              min(dp[i + 1][j + 1][k], dp[i][j][k] + (s[i + 1] != t[j]));
        }
        dp[i + 1][nxt[j]][k] =
            min(dp[i + 1][nxt[j]][k], dp[i][j][k] + (s[i + 1] == t[j]));
      }
    }
  }
  for (long long i = 0; i < n - m + 2; i++) {
    long long mn = inf;
    for (long long j = 0; j < m; j++) {
      mn = min(mn, dp[n][j][i]);
    }
    if (mn == inf) mn = -1;
    cout << mn << ' ';
  }
}

### 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 long long inf = 2e9;
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  long long n, m;
  string s, t;
  cin >> n >> m >> s >> t;
  vector<long long> pi(m, 0), nxt(m, 0);
  for (long long i = 1; i < m; i++) {
    {
      t[i] = char('0' + (t[i] - '0') ^ 1);
      long long j = pi[i - 1];
      while (j > 0 and t[i] != t[j]) {
        j = pi[j - 1];
      }
      if (t[i] == t[j]) j++;
      nxt[i] = j;
      t[i] = char('0' + (t[i] - '0') ^ 1);
    }
    {
      long long j = pi[i - 1];
      while (j > 0 and t[i] != t[j]) {
        j = pi[j - 1];
      }
      if (t[i] == t[j]) j++;
      pi[i] = j;
    }
  }
  s = '#' + s;
  long long dp[n + 1][m][n - m + 2];
  fill(**dp, **dp + sizeof dp / sizeof(long long), inf);
  dp[0][0][0] = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      for (long long k = 0; k < n - m + 2; k++) {
        if (dp[i][j][k] == inf) continue;
        if (j == m - 1) {
          dp[i + 1][pi.back()][k + 1] = min(dp[i + 1][pi.back()][k + 1],
                                            dp[i][j][k] + (s[i + 1] != t[j]));
        } else {
          dp[i + 1][j + 1][k] =
              min(dp[i + 1][j + 1][k], dp[i][j][k] + (s[i + 1] != t[j]));
        }
        dp[i + 1][nxt[j]][k] =
            min(dp[i + 1][nxt[j]][k], dp[i][j][k] + (s[i + 1] == t[j]));
      }
    }
  }
  for (long long i = 0; i < n - m + 2; i++) {
    long long mn = inf;
    for (long long j = 0; j < m; j++) {
      mn = min(mn, dp[n][j][i]);
    }
    if (mn == inf) mn = -1;
    cout << mn << ' ';
  }
}

```