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 n, len;
char s[2010], c[2010];
int f[2010][2010];
int cal(int x) {
  if (x < len) return 0x7f7f7f7f;
  int ans = 0, p = len, q = x;
  while (p && q) {
    if (s[q] == c[p])
      p--, q--;
    else
      ans++, q--;
  }
  if (p == 0) return ans;
  return 0x7f7f7f7f;
}
int main() {
  scanf("%s%s", s + 1, c + 1);
  n = strlen(s + 1), len = strlen(c + 1);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++)
      if (i < j) f[i][j] = -0x7f7f7f7f;
  for (int i = 1; i <= n; i++) {
    int x = cal(i);
    for (int j = 0; j <= n; j++) {
      f[i][j] = max(f[i][j], f[i - 1][j]);
      if (j - x >= 0) f[i][j] = max(f[i][j], f[i - len - x][j - x] + 1);
    }
  }
  for (int i = 0; i <= n; i++) printf("%d ", f[n][i]);
  return 0;
}
