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 s[2005], p[506];
int F[2005][2005], n, m;
int main() {
  int i, j, k;
  scanf("%s", s + 1);
  scanf("%s", p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  for (i = 1; i <= n; i++) {
    j = i, k = m;
    while (j && k) {
      if (s[j] == p[k]) k--;
      j--;
    }
    if (!k) {
      for (k = 0; k <= i; k++)
        if (k >= i - j - m && k - i + m <= 0) F[i][k] = F[j][k - i + j + m] + 1;
    }
    for (k = 0; k <= i; k++) F[i][k] = max(F[i][k], F[i - 1][k]);
  }
  for (i = 0; i <= n; i++) printf("%d ", F[n][i]);
  return 0;
}
