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;
const int inf = 1000000;
int d[2100], f[600][2100], h[600][2100];
char c1[2100], c2[1200];
int i, j, k, n1, n2;
int main() {
  scanf("%s", c1 + 1);
  scanf("%s", c2 + 1);
  n1 = strlen(c1 + 1);
  n2 = strlen(c2 + 1);
  memset(f, 250, sizeof(f));
  f[n2][0] = 0;
  for (i = 1; i <= n1; i++) {
    for (j = 1; j <= n2; j++)
      for (k = 0; k <= n1 / n2; k++) h[j][k] = f[j][k], f[j][k] = -inf;
    for (j = 1; j <= n2; j++)
      for (k = 0; k <= n1 / n2; k++) {
        if (c1[i] == c2[j])
          if ((j == 1) && (k > 0))
            f[j][k] = h[n2][k - 1] + 1;
          else
            f[j][k] = h[j - 1][k] + 1;
        if (j < n2)
          f[j][k] = max(f[j][k], h[j][k]);
        else
          f[j][k] = max(f[j][k], h[j][k] + 1);
      }
  }
  for (k = 1; k <= n1 / n2; k++)
    if (f[n2][k] >= 0) d[n1 - f[n2][k]] = k;
  for (k = 1; k <= n1; k++) d[k] = max(d[k], d[k - 1]);
  for (k = 0; k <= n1; k++) printf("%d ", min(d[k], (n1 - k) / n2));
  return 0;
}
