#include <bits/stdc++.h>
using namespace std;
int N, M;
string A, B;
int to[500], ed;
int dp[2][505][501];
int main() {
  cin >> N >> M >> A >> B;
  for (int i = 0; i < M; i++) {
    string now = B.substr(0, i);
    now += B[i] == '0' ? '1' : '0';
    for (int k = 0; k <= now.size(); k++) {
      if (now.substr(k) == B.substr(0, now.size() - k)) {
        to[i] = now.size() - k;
        break;
      }
    }
  }
  for (int k = 1; k <= M; k++)
    if (B.substr(k) == B.substr(0, M - k)) {
      ed = M - k;
      break;
    }
  int now = 0;
  int mx = N - M + 1;
  for (int i = 0; i <= mx; i++)
    for (int j = 0; j < M; j++) dp[now][i][j] = 1e9;
  dp[now][0][0] = 0;
  for (int i = 0; i < N; i++) {
    int nxt = 1 - now;
    for (int j = 0; j <= mx; j++)
      for (int k = 0; k < M; k++) dp[nxt][j][k] = 1e9;
    for (int j = 0; j <= mx; j++)
      for (int k = 0; k < M; k++) {
        if (A[i] == B[k]) {
          if (k + 1 == M)
            dp[nxt][j + 1][ed] = min(dp[nxt][j + 1][ed], dp[now][j][k]);
          else
            dp[nxt][j][k + 1] = min(dp[nxt][j][k + 1], dp[now][j][k]);
          dp[nxt][j][to[k]] = min(dp[nxt][j][to[k]], dp[now][j][k] + 1);
        } else {
          if (k + 1 == M)
            dp[nxt][j + 1][ed] = min(dp[nxt][j + 1][ed], dp[now][j][k] + 1);
          else
            dp[nxt][j][k + 1] = min(dp[nxt][j][k + 1], dp[now][j][k] + 1);
          dp[nxt][j][to[k]] = min(dp[nxt][j][to[k]], dp[now][j][k]);
        }
      }
    now = nxt;
  }
  for (int k = 0; k <= mx; k++) {
    int ans = 1e9;
    for (int j = 0; j < M; j++) ans = min(ans, dp[now][k][j]);
    if (ans == (int)1e9) ans = -1;
    cout << ans << (k == mx ? "\n" : " ");
  }
}

### 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>
using namespace std;
int N, M;
string A, B;
int to[500], ed;
int dp[2][505][501];
int main() {
  cin >> N >> M >> A >> B;
  for (int i = 0; i < M; i++) {
    string now = B.substr(0, i);
    now += B[i] == '0' ? '1' : '0';
    for (int k = 0; k <= now.size(); k++) {
      if (now.substr(k) == B.substr(0, now.size() - k)) {
        to[i] = now.size() - k;
        break;
      }
    }
  }
  for (int k = 1; k <= M; k++)
    if (B.substr(k) == B.substr(0, M - k)) {
      ed = M - k;
      break;
    }
  int now = 0;
  int mx = N - M + 1;
  for (int i = 0; i <= mx; i++)
    for (int j = 0; j < M; j++) dp[now][i][j] = 1e9;
  dp[now][0][0] = 0;
  for (int i = 0; i < N; i++) {
    int nxt = 1 - now;
    for (int j = 0; j <= mx; j++)
      for (int k = 0; k < M; k++) dp[nxt][j][k] = 1e9;
    for (int j = 0; j <= mx; j++)
      for (int k = 0; k < M; k++) {
        if (A[i] == B[k]) {
          if (k + 1 == M)
            dp[nxt][j + 1][ed] = min(dp[nxt][j + 1][ed], dp[now][j][k]);
          else
            dp[nxt][j][k + 1] = min(dp[nxt][j][k + 1], dp[now][j][k]);
          dp[nxt][j][to[k]] = min(dp[nxt][j][to[k]], dp[now][j][k] + 1);
        } else {
          if (k + 1 == M)
            dp[nxt][j + 1][ed] = min(dp[nxt][j + 1][ed], dp[now][j][k] + 1);
          else
            dp[nxt][j][k + 1] = min(dp[nxt][j][k + 1], dp[now][j][k] + 1);
          dp[nxt][j][to[k]] = min(dp[nxt][j][to[k]], dp[now][j][k]);
        }
      }
    now = nxt;
  }
  for (int k = 0; k <= mx; k++) {
    int ans = 1e9;
    for (int j = 0; j < M; j++) ans = min(ans, dp[now][k][j]);
    if (ans == (int)1e9) ans = -1;
    cout << ans << (k == mx ? "\n" : " ");
  }
}

```