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 = 505;
int cost[N][N];
vector<vector<vector<int>>> dp;
string s, t;
int n, m;
int pf[N];
int pfc[N];
int z_function(string s) {
  int n = (int)s.length();
  vector<int> z(n);
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z[n - 1];
}
int K;
const int inf = 1e9 + 5;
int solve(int i, int j, int k) {
  if (k < 0) return inf;
  if (i == n) {
    if (k) return inf;
    return 0;
  }
  int &res = dp[i][j][k];
  if (~res) return res;
  res = inf;
  int c = 0;
  if (s[i] != t[j]) c++;
  int dc = c ^ 1;
  {
    int nj = j + 1;
    int nk = k;
    if (nj == m) {
      nk = k - 1;
      nj = pf[m - 1];
    }
    res = min(res, c + solve(i + 1, nj, nk));
    res = min(res, dc + solve(i + 1, pfc[j], k));
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  cin >> s >> t;
  K = n - m + 2;
  dp = vector<vector<vector<int>>>(
      n + 1, vector<vector<int>>(m + 1, vector<int>(n - m + 3)));
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= m; j++) {
      for (int k = 0; k < K; k++) dp[i][j][k] = -1;
    }
  }
  pf[m - 1] = z_function(t);
  for (int i = 0; i < m; i++) {
    string temp;
    for (int j = 0; j < i; j++) temp += t[j];
    if (t[i] == '0')
      temp += '1';
    else
      temp += '0';
    pfc[i] = z_function(temp);
  }
  for (int k = 0; k < K; k++) {
    int ans = solve(0, 0, k);
    if (ans >= inf)
      cout << -1 << " ";
    else
      cout << ans << " ";
  }
}
