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;
char s[2005];
char t[505];
int w[2005];
int p[2005];
int f[2005][2005];
int main() {
  scanf("%s %s", s + 1, t + 1);
  int n = strlen(s + 1);
  int m = strlen(t + 1);
  for (int i = 1; i <= n; ++i) {
    w[i] = 1e9;
    for (int j = i, k = m; j >= 1; --j) {
      if (s[j] == t[k]) {
        --k;
      }
      if (k == 0) {
        w[i] = (i - j + 1) - m;
        p[i] = j - 1;
        break;
      }
    }
  }
  for (int i = 0; i <= n; ++i) {
    f[0][i] = -1e9;
  }
  f[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= n; ++j) {
      f[i][j] = f[i - 1][j];
    }
    for (int j = 0; j <= n; ++j) {
      if (j + 1 <= n) {
        if (f[i - 1][j] > f[i][j + 1]) {
          f[i][j + 1] = f[i - 1][j];
        }
      }
      if (j + w[i] <= n) {
        if (f[p[i]][j] + 1 > f[i][j + w[i]]) {
          f[i][j + w[i]] = f[p[i]][j] + 1;
        }
      }
    }
  }
  for (int i = 0; i <= n; ++i) {
    printf("%d ", f[n][i]);
  }
  return 0;
}
