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 maxn = 505;
int f[maxn], nxt[maxn][2];
void getf(string s) {
  f[0] = 0;
  for (int i = 1; i < s.size(); ++i) {
    int j = f[i - 1];
    while (j && s[i] != s[j]) j = f[j - 1];
    if (s[i] == s[j])
      f[i] = j + 1;
    else
      f[i] = 0;
  }
  for (int i = 0; i < s.size(); ++i) {
    for (int j = 0; j < 2; ++j) {
      int tmp = i;
      while (tmp && s[i] - '0' != j) tmp = f[tmp - 1];
      if (s[tmp] - '0' == j)
        nxt[i][j] = tmp + 1;
      else
        nxt[i][j] = 0;
    }
  }
}
int dp[505][505][505];
void cmin(int &a, int b) { a = min(a, b); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  string s, t;
  cin >> s >> t;
  getf(t);
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0][0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= i; ++j) {
      for (int k = 0; k < m; ++k) {
        if (dp[i][j][k] >= 1e9) continue;
        cmin(dp[i + 1][j + (nxt[k][0] == m)]
               [(nxt[k][0] == m ? f[k] : nxt[k][0])],
             dp[i][j][k] + (s[i] != '0'));
        cmin(dp[i + 1][j + (nxt[k][1] == m)]
               [(nxt[k][1] == m ? f[k] : nxt[k][1])],
             dp[i][j][k] + (s[i] != '1'));
      }
    }
  }
  vector<int> ans(n + 1, 1e9);
  for (int j = 0; j <= n; ++j) {
    for (int k = 0; k < t.size(); ++k) {
      ans[j] = min(ans[j], dp[n][j][k]);
    }
  }
  for (int i = 0; i < n - m + 2; ++i) {
    if (ans[i] >= 1e9)
      cout << -1 << " ";
    else
      cout << ans[i] << " ";
  }
  cout << endl;
}
