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.
#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
const int MM = 998244353;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
const int N = 502;
int dp[N][N][N];
int f[N][2];
int _runtimeTerror_() {
  int n, m;
  cin >> n >> m;
  string a, b;
  cin >> a >> b;
  vector<int> pi(m + 1, 0);
  for (int i = 1; i < m; ++i) {
    int j = pi[i - 1];
    while (j > 0 && b[i] != b[j]) {
      j = pi[j - 1];
    }
    if (b[i] == b[j]) {
      ++j;
    } else {
      j = 0;
    }
    pi[i] = j;
  }
  f[0][0] = b[0] == '0';
  f[0][1] = b[0] == '1';
  for (int i = 1; i < m; ++i) {
    int j = i;
    while (j > 0 && '0' != b[j]) {
      j = pi[j - 1];
    }
    if (b[j] == '0') {
      ++j;
    } else {
      j = 0;
    }
    f[i][0] = j;
    j = i;
    while (j > 0 && '1' != b[j]) {
      j = pi[j - 1];
    }
    if (b[j] == '1') {
      ++j;
    } else {
      j = 0;
    }
    f[i][1] = j;
  }
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= m; ++j) {
      for (int k = 0; k <= n - m + 1; ++k) {
        dp[i][j][k] = 10000000;
      }
    }
  }
  dp[0][0][0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      for (int k = 0; k <= n - m + 1; ++k) {
        if (j == m - 1) {
          int z = f[j][0];
          if (z == m) {
            amin(dp[i + 1][pi[m - 1]][k + 1], dp[i][j][k] + (a[i] != '0'));
          } else {
            amin(dp[i + 1][f[j][0]][k], dp[i][j][k] + (a[i] != '0'));
          }
          z = f[j][1];
          if (z == m) {
            amin(dp[i + 1][pi[m - 1]][k + 1], dp[i][j][k] + (a[i] != '1'));
          } else {
            amin(dp[i + 1][f[j][0]][k], dp[i][j][k] + (a[i] != '1'));
          }
        } else {
          amin(dp[i + 1][f[j][0]][k], dp[i][j][k] + (a[i] != '0'));
          amin(dp[i + 1][f[j][1]][k], dp[i][j][k] + (a[i] != '1'));
        }
      }
    }
  }
  for (int i = 0; i <= n - m + 1; ++i) {
    long long ans = 10000000;
    for (int j = 0; j < m; ++j) {
      amin(ans, dp[n][j][i]);
    }
    if (ans >= 1000) {
      ans = -1;
    }
    cout << ans << " ";
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  return 0;
}
