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 lim = 2e3 + 10, inf = 2e9;
char a[lim], b[lim];
int f[lim][lim];
int pos[lim], res[lim];
int ans[lim];
int main() {
  scanf("%s", a + 1);
  scanf("%s", b + 1);
  int n = strlen(a + 1), m = strlen(b + 1);
  for (int i = 1; i <= n; i++)
    if (a[i] == b[1]) {
      int k = 1;
      for (int j = i; j <= n; j++) {
        if (a[j] == b[k]) {
          k++;
          if (k == m + 1) {
            pos[i] = j;
            break;
          }
        } else
          res[i]++;
      }
    }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      f[i][j] = -inf;
    }
  }
  f[0][0] = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++)
      if (f[i][j] != -inf) {
        f[i + 1][j + 1] = max(f[i + 1][j + 1], f[i][j]);
        f[i + 1][j] = max(f[i + 1][j], f[i][j]);
        if (pos[i + 1]) {
          f[pos[i + 1]][j + res[i + 1]] =
              max(f[pos[i + 1]][j + res[i + 1]], f[i][j] + 1);
        }
      }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) ans[j] = max(ans[j], f[i][j]);
  }
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  printf("%d\n", ans[n]);
  return 0;
}
