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[2010], p[2010];
int n, m, i, j, now;
int f[2010][2010];
int End[2010];
inline int max(int a, int b) {
  if (a > b)
    return (a);
  else
    return (b);
}
int main() {
  scanf("%s", s + 1);
  scanf("%s", p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  for (i = 1; i <= n; ++i) {
    now = 1;
    for (j = i; j <= n && now <= m; ++j)
      if (s[j] == p[now]) ++now;
    if (now > m) End[i] = j;
  }
  for (i = 0; i < n; ++i)
    for (j = 0; j <= i; ++j) {
      f[i + 1][j] = max(f[i + 1][j], f[i][j]);
      f[i + 1][j + 1] = max(f[i + 1][j + 1], f[i][j]);
      if (End[i + 1])
        f[End[i + 1] - 1][j + End[i + 1] - i - 1 - m] =
            max(f[End[i + 1] - 1][j + End[i + 1] - i - 1 - m], f[i][j] + 1);
    }
  for (i = 0; i < n; ++i) printf("%d ", f[n][i]);
  printf("%d\n", f[n][n]);
  return 0;
}
