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 N = 2005;
int n, m;
char s[N], t[N];
int pre[N], val[N];
int f[N][N];
void work() {
  int i, j, k, x;
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  for (i = 1; i <= n; ++i) {
    x = 0;
    for (j = i, k = m; j > 0 && k; --j) {
      if (s[j] == t[k])
        --k;
      else
        ++x;
    }
    if (k == 0) {
      pre[i] = j;
      val[i] = x;
    } else {
      pre[i] = -1;
    }
  }
  memset(f, 0xC0, sizeof(f));
  f[0][0] = 0;
  for (i = 1; i <= n; ++i) {
    for (j = 0; j <= i; ++j) {
      f[i][j] = max(f[i][j], f[i - 1][j]);
      if (j) f[i][j] = max(f[i][j], f[i - 1][j - 1]);
      if (~pre[i] && j >= val[i])
        f[i][j] = max(f[i][j], f[pre[i]][j - val[i]] + 1);
    }
  }
  for (i = 0; i <= n; ++i) printf("%d%c", f[n][i], " \n"[i == n]);
}
int main() {
  work();
  return 0;
}
