#include <bits/stdc++.h>
using namespace std;
const int maxn = 505;
inline void down(int &a, const int &b) {
  if (a == -1 || a > b) a = b;
}
int n, m;
char A[maxn], B[maxn];
int f[2][maxn][maxn];
int ans[maxn];
struct AC {
  int root, tot;
  int son[maxn][2], fail[maxn], ed[maxn];
  int newnode() {
    ++tot;
    memset(son[tot], 0, sizeof son[tot]);
    fail[tot] = 0;
    ed[tot] = 0;
    return tot;
  }
  void init() {
    tot = 0;
    root = newnode();
  }
  queue<int> q;
  void build(char str[], int len) {
    int x = root;
    for (int i = 0; i < len; i++) {
      int c = str[i] - '0';
      if (!son[x][c]) son[x][c] = newnode();
      x = son[x][c];
      if (i == len - 1) ed[x] = 1;
    }
    q.push(root);
    while (!q.empty()) {
      x = q.front();
      q.pop();
      int fl = fail[x];
      if (ed[fl]) ed[x] = 1;
      for (int i = 0; i < 2; i++) {
        int &y = son[x][i];
        if (y) {
          if (fl)
            fail[y] = son[fl][i];
          else
            fail[y] = root;
          q.push(y);
        } else {
          if (fl)
            y = son[fl][i];
          else
            y = root;
        }
      }
    }
  }
  void dp(char str[], int len, int uj) {
    memset(f, -1, sizeof f);
    int now = 0;
    f[now][root][0] = 0;
    for (int i = 0; i < len; i++) {
      now = !now;
      int cc = str[i] - '0';
      for (int x = 1; x <= tot; x++)
        for (int j = 0; j <= uj; j++)
          if (f[!now][x][j] != -1) {
            int &temp = f[!now][x][j];
            int y = son[x][cc];
            down(f[now][y][j + ed[y]], temp + 0);
            y = son[x][!cc];
            down(f[now][y][j + ed[y]], temp + 1);
            temp = -1;
          }
    }
    for (int j = 0; j <= uj; j++) ans[j] = -1;
    for (int x = 1; x <= tot; x++)
      for (int j = 0; j <= uj; j++)
        if (f[now][x][j] != -1) {
          down(ans[j], f[now][x][j]);
        }
  }
} ac;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  cin >> (A) >> (B);
  ac.init();
  ac.build(B, m);
  ac.dp(A, n, n - m + 1);
  for (int i = 0; i <= n - m + 1; i++) cout << ans[i] << " \n"[i == n - m + 1];
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 505;
inline void down(int &a, const int &b) {
  if (a == -1 || a > b) a = b;
}
int n, m;
char A[maxn], B[maxn];
int f[2][maxn][maxn];
int ans[maxn];
struct AC {
  int root, tot;
  int son[maxn][2], fail[maxn], ed[maxn];
  int newnode() {
    ++tot;
    memset(son[tot], 0, sizeof son[tot]);
    fail[tot] = 0;
    ed[tot] = 0;
    return tot;
  }
  void init() {
    tot = 0;
    root = newnode();
  }
  queue<int> q;
  void build(char str[], int len) {
    int x = root;
    for (int i = 0; i < len; i++) {
      int c = str[i] - '0';
      if (!son[x][c]) son[x][c] = newnode();
      x = son[x][c];
      if (i == len - 1) ed[x] = 1;
    }
    q.push(root);
    while (!q.empty()) {
      x = q.front();
      q.pop();
      int fl = fail[x];
      if (ed[fl]) ed[x] = 1;
      for (int i = 0; i < 2; i++) {
        int &y = son[x][i];
        if (y) {
          if (fl)
            fail[y] = son[fl][i];
          else
            fail[y] = root;
          q.push(y);
        } else {
          if (fl)
            y = son[fl][i];
          else
            y = root;
        }
      }
    }
  }
  void dp(char str[], int len, int uj) {
    memset(f, -1, sizeof f);
    int now = 0;
    f[now][root][0] = 0;
    for (int i = 0; i < len; i++) {
      now = !now;
      int cc = str[i] - '0';
      for (int x = 1; x <= tot; x++)
        for (int j = 0; j <= uj; j++)
          if (f[!now][x][j] != -1) {
            int &temp = f[!now][x][j];
            int y = son[x][cc];
            down(f[now][y][j + ed[y]], temp + 0);
            y = son[x][!cc];
            down(f[now][y][j + ed[y]], temp + 1);
            temp = -1;
          }
    }
    for (int j = 0; j <= uj; j++) ans[j] = -1;
    for (int x = 1; x <= tot; x++)
      for (int j = 0; j <= uj; j++)
        if (f[now][x][j] != -1) {
          down(ans[j], f[now][x][j]);
        }
  }
} ac;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  cin >> (A) >> (B);
  ac.init();
  ac.build(B, m);
  ac.dp(A, n, n - m + 1);
  for (int i = 0; i <= n - m + 1; i++) cout << ans[i] << " \n"[i == n - m + 1];
  return 0;
}

```