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>
char s[2000 + 20], t[2000 + 20];
int pre[2000 + 20], ans[2000 + 20];
int dp[2000 + 20][2000 + 20];
int main() {
  scanf("%s%s", s + 1, t + 1);
  int n = strlen(s + 1);
  int m = strlen(t + 1);
  for (int i = 1; i <= n; i++) {
    if (s[i] == t[1]) {
      for (int j = i, k = 1; j <= n; j++) {
        if (s[j] == t[k]) k++;
        if (k > m) {
          pre[j] = i;
          break;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!pre[i]) pre[i] = pre[i - 1];
  }
  for (int i = 1; i <= n; i++) {
    dp[0][i] = 20000;
  }
  for (int j = 1; j <= n; j++) {
    for (int i = 1; i <= n; i++) {
      dp[i][j] = dp[i - 1][j];
      if (!pre[i]) continue;
      dp[i][j] = std::min(dp[i][j], dp[pre[i] - 1][j - 1] + i + 1 - pre[i] - m);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = dp[i][j]; k <= i - j * m; k++) ans[k] = std::max(ans[k], j);
    }
  }
  for (int i = 0; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}
