Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a result. Then he calculates <image> that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s'. He wants to make this number as big as possible.

More formally, let's define <image> as maximum value of <image> over all s' that can be obtained by removing exactly x characters from s. Dreamoon wants to know <image> for all x from 0 to |s| where |s| denotes the length of string s.

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 2 000).

The second line of the input contains the string p (1 ≤ |p| ≤ 500).

Both strings will only consist of lower case English letters.

Output

Print |s| + 1 space-separated integers in a single line representing the <image> for all x from 0 to |s|.

Examples

Input

aaaaa
aa


Output

2 2 1 1 0 0


Input

axbaxxb
ab


Output

0 1 1 2 1 1 0 0

Note

For the first sample, the corresponding optimal values of s' after removal 0 through |s| = 5 characters from s are {"aaaaa", "aaaa", "aaa", "aa", "a", ""}. 

For the second sample, possible corresponding optimal values of s' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab", "a", ""}.
#include <bits/stdc++.h>
using namespace std;
int main() {
  string s, p;
  cin >> s >> p;
  int n = s.size(), l = p.size();
  if (n < l) {
    for (int i = 0; i < n; i++) cout << 0 << " ";
    cout << 0 << endl;
    return 0;
  }
  int m = n / l + 1;
  int dp[n + 1][m];
  int inf = 100000000;
  fill_n(dp[0], (n + 1) * m, inf);
  for (int i = 0; i <= n; i++) {
    dp[i][0] = 0;
  }
  for (int i = n - l; i >= 0; i--) {
    if (s[i] == p[0]) {
      int index = i;
      int rem = 0;
      for (int j = 0; j < l; j++) {
        while (index < n && s[index] != p[j]) {
          index++;
          rem++;
        }
        index++;
        if (index > n) break;
      }
      if (index > n) continue;
      for (int j = 1; j < m; j++)
        dp[i][j] = min(dp[i + 1][j], rem + dp[index][j - 1]);
    } else {
      for (int j = 0; j < m; j++) {
        dp[i][j] = dp[i + 1][j];
      }
    }
  }
  vector<int> res(n + 1, 0);
  for (int k = 1; k < m; k++) {
    int mini = inf;
    for (int i = 0; i < n; i++) {
      mini = min(mini, dp[i][k]);
    }
    for (int i = mini; i <= n - k * l; i++) res[i] = k;
  }
  cout << res[0];
  for (int i = 1; i <= n; i++) cout << " " << res[i];
  cout << endl;
}
