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 A[2100];
int D[2100][2100];
int main() {
  setvbuf(stdout, NULL, _IONBF, 0);
  char c[2100];
  scanf("%s", c);
  string s(c);
  char d[600];
  scanf("%s", d);
  string p(d);
  for (int i = 0; i < s.size(); i++) {
    int ps = i;
    int pp;
    for (pp = 0; pp < p.size(); pp++) {
      while (ps < s.size() && p[pp] != s[ps]) {
        ps++;
      }
      if (ps == s.size()) {
        break;
      } else {
        ps++;
      }
    }
    if (pp == p.size()) {
      A[i] = ps - i;
    } else {
      A[i] = -1;
    }
  }
  for (int i = 0; i < s.size(); i++) {
    for (int r = 0; r <= i; r++) {
      D[i + 1][r] = max(D[i + 1][r], D[i][r]);
      D[i + 1][r + 1] = max(D[i + 1][r + 1], D[i][r]);
      if (A[i] >= 0) {
        D[i + A[i]][r + A[i] - p.size()] =
            max(D[i + A[i]][r + A[i] - p.size()], D[i][r] + 1);
      }
    }
  }
  printf("%d", D[s.size()][0]);
  for (int i = 1; i <= s.size(); i++) {
    printf(" %d", D[s.size()][i]);
  }
  printf("\n");
}
