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;
char a[2222], b[2222];
int n1, n2, i, j, k, p[2222], dp[2222][2222], mm;
int main() {
  gets(a);
  gets(b);
  n1 = strlen(a);
  n2 = strlen(b);
  for (i = n1 - 1; i >= 0; i--) {
    if (a[i] == b[n2 - 1]) {
      j = i + 1;
      for (k = n2 - 1; k >= 0 && j >= 0; k--) {
        j--;
        while (a[j] != b[k] && j >= 0) j--;
      }
      if (k == -1 && a[j] == b[0])
        p[i] = j;
      else
        p[i] = -1;
    } else
      p[i] = -1;
  }
  for (i = 0; i < n1; i++) {
    for (j = 0; j <= i + 1; j++) {
      dp[i][j] = max(i == 0 ? 0 : dp[i - 1][j],
                     p[i] == -1 || n2 + j < i - p[i] + 1 ? 0
                     : p[i] == 0                         ? 1
                                 : dp[p[i] - 1][n2 - (i - p[i] + 1) + j] + 1);
    }
  }
  mm = 0;
  for (i = 0; i <= n1; i++) {
    mm = max(mm, dp[n1 - 1][i]);
    mm = min(mm, (n1 - i) / n2);
    printf("%d ", mm);
  }
  return 0;
}
