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_MAX = 500;
const int M_MAX = 500;
const int INF = INT_MAX / 2;
int n, m;
string a, b;
int kmp[M_MAX];
int prec[M_MAX + 2][2];
int dp[N_MAX][N_MAX + 2][M_MAX + 2];
void update(int &x, int y) { x = min(x, y); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  cin >> a >> b;
  kmp[0] = 0;
  for (int i = 1; i < m; i++) {
    kmp[i] = kmp[i - 1];
    while (kmp[i] > 0 && b[kmp[i]] != b[i]) kmp[i] = kmp[kmp[i] - 1];
    if (b[kmp[i]] == b[i]) kmp[i]++;
  }
  for (int i = 0; i < n; i++)
    for (int cnt = 0; cnt <= n; cnt++)
      for (int pref = 0; pref <= m; pref++) dp[i][cnt][pref] = INF;
  dp[0][(m == 1)][1] = (a[0] != b[0]);
  dp[0][0][0] = 0;
  for (int pref = 0; pref <= m; pref++) {
    for (char c = '0'; c <= '1'; c++) {
      int newPref = (pref == m ? kmp[pref - 1] : pref);
      while (newPref > 0 && b[newPref] != c) newPref = kmp[newPref - 1];
      if (b[newPref] == c) newPref++;
      prec[pref][c - '0'] = newPref;
    }
  }
  for (int i = 0; i + 1 < n; i++)
    for (int cnt = 0; cnt <= n; cnt++)
      for (int pref = 0; pref <= m; pref++) {
        for (char c = '0'; c <= '1'; c++) {
          int newPref = prec[pref][c - '0'];
          update(dp[i + 1][cnt + (newPref == m)][newPref],
                 dp[i][cnt][pref] + (a[i + 1] != c));
        }
      }
  for (int cnt = 0; cnt <= n - m + 1; cnt++) {
    int mn = INF;
    for (int pref = 0; pref <= m; pref++) mn = min(mn, dp[n - 1][cnt][pref]);
    cout << (mn == INF ? -1 : mn) << " ";
  }
  cout << "\n";
  return 0;
}
