#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
void SolveTask() {
  int n, m;
  string a, b;
  cin >> n >> m >> a >> b;
  vector<vector<int>> pf(m + 1, vector<int>(2, 0));
  for (int i = 0; i <= m; ++i) {
    for (int r = min(i, m - 1); r >= 0; --r) {
      bool ok = true;
      for (int j = r - 1, h = i - 1; j >= 0; --j, --h) {
        if (b[j] != b[h]) {
          ok = false;
          break;
        }
      }
      if (ok) {
        pf[i][b[r] - '0'] = max(pf[i][b[r] - '0'], r + 1);
      }
    }
  }
  const auto SetMin = [](int& x, int y) { x = min(x, y); };
  vector<vector<vector<int>>> dp(
      n + 1, vector<vector<int>>(n + 2, vector<int>(m + 1, INF)));
  dp[0][0][0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= max(0, i - m + 2); ++j) {
      for (int k = 0; k <= min(i, m); ++k) {
        if (dp[i][j][k] == INF) {
          continue;
        }
        const int c = a[i] - '0';
        SetMin(dp[i + 1][j + (pf[k][c] == m)][pf[k][c]], dp[i][j][k]);
        SetMin(dp[i + 1][j + (pf[k][c ^ 1] == m)][pf[k][c ^ 1]],
               dp[i][j][k] + 1);
      }
    }
  }
  vector<int> ans(n - m + 2, INF);
  for (int p = 0; p <= m; ++p) {
    for (int j = 0; j <= n - m + 2; ++j) {
      if (dp[n][j][p] != INF) {
        SetMin(ans[j], dp[n][j][p]);
      }
    }
  }
  for (int x : ans) {
    x = x == INF ? -1 : x;
    cout << x << ' ';
  }
  cout << "\n";
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int test_cases = 1;
  for (int yt = 1; yt <= test_cases; ++yt) {
    SolveTask();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
using ll = long long;
using pii = pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
void SolveTask() {
  int n, m;
  string a, b;
  cin >> n >> m >> a >> b;
  vector<vector<int>> pf(m + 1, vector<int>(2, 0));
  for (int i = 0; i <= m; ++i) {
    for (int r = min(i, m - 1); r >= 0; --r) {
      bool ok = true;
      for (int j = r - 1, h = i - 1; j >= 0; --j, --h) {
        if (b[j] != b[h]) {
          ok = false;
          break;
        }
      }
      if (ok) {
        pf[i][b[r] - '0'] = max(pf[i][b[r] - '0'], r + 1);
      }
    }
  }
  const auto SetMin = [](int& x, int y) { x = min(x, y); };
  vector<vector<vector<int>>> dp(
      n + 1, vector<vector<int>>(n + 2, vector<int>(m + 1, INF)));
  dp[0][0][0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= max(0, i - m + 2); ++j) {
      for (int k = 0; k <= min(i, m); ++k) {
        if (dp[i][j][k] == INF) {
          continue;
        }
        const int c = a[i] - '0';
        SetMin(dp[i + 1][j + (pf[k][c] == m)][pf[k][c]], dp[i][j][k]);
        SetMin(dp[i + 1][j + (pf[k][c ^ 1] == m)][pf[k][c ^ 1]],
               dp[i][j][k] + 1);
      }
    }
  }
  vector<int> ans(n - m + 2, INF);
  for (int p = 0; p <= m; ++p) {
    for (int j = 0; j <= n - m + 2; ++j) {
      if (dp[n][j][p] != INF) {
        SetMin(ans[j], dp[n][j][p]);
      }
    }
  }
  for (int x : ans) {
    x = x == INF ? -1 : x;
    cout << x << ' ';
  }
  cout << "\n";
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int test_cases = 1;
  for (int yt = 1; yt <= test_cases; ++yt) {
    SolveTask();
  }
  return 0;
}

```