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 N = 502;
int f[N][N][N], nxt[N];
int n, m, q, i, j, k, x;
char s[N], t[N];
void getmn(int &x, int y) {
  if (x > y) x = y;
}
void get_next(char *s, int *nxt) {
  int n, i, j;
  n = strlen(s + 1);
  j = 0;
  nxt[1] = 0;
  for (i = 2; i <= n; i++) {
    while (j && s[i] != s[j + 1]) j = nxt[j];
    j += s[i] == s[j + 1];
    nxt[i] = j;
  }
}
int main() {
  cin >> n >> m >> s + 1 >> t + 1;
  get_next(t, nxt);
  memset(f, 0x3f, sizeof f);
  int inf = f[0][0][0];
  f[0][0][0] = 0;
  for (i = 0; i <= n; i++)
    for (j = 0; j <= n; j++)
      for (k = 0; k < m; k++)
        if (f[i][j][k] != inf) {
          if (s[i + 1] == t[k + 1]) {
            if (k == m - 1)
              getmn(f[i + 1][j + 1][nxt[m]], f[i][j][k]);
            else
              getmn(f[i + 1][j][k + 1], f[i][j][k]);
            x = nxt[k];
            getmn(f[i + 1][j][x + ((s[i + 1] ^ 1) == t[x + 1])],
                  f[i][j][k] + 1);
          } else {
            if (k == m - 1)
              getmn(f[i + 1][j + 1][nxt[m]], f[i][j][k] + 1);
            else
              getmn(f[i + 1][j][k + 1], f[i][j][k] + 1);
            x = nxt[k];
            getmn(f[i + 1][j][x + (s[i + 1] == t[x + 1])], f[i][j][k]);
          }
        }
  for (i = 0; i <= n - m + 1; i++) {
    x = *min_element(f[n][i], f[n][i] + m);
    cout << (x == inf ? -1 : x) << ' ';
  }
}
