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 const N = 2111;
int f[N][N], o[N + 1];
char s[N], p[N];
int main() {
  scanf("%s%s", s, p);
  int ls = strlen(s), lp = strlen(p);
  for (int i = 0; i < ls; ++i)
    for (int j = 0; j <= i; ++j) {
      int k = j % lp, l = (j + 1) / lp, &x = f[i + 1][j + 1] = N,
          y = i + 1 - l * lp;
      if (i >= j + 1) x = f[i][j + 1] + !!(k + 1 - lp);
      if (s[i] == p[k]) x = min(x, f[i][j]);
      if (i + 1 == ls)
        for (; y >= x; --y) o[y] = max(o[y], l);
    }
  for (int i = 0; i <= ls; ++i) printf("%d ", o[i]);
  return 0;
}
