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], p[505];
int A[2005];
int D[2005][2005];
int main() {
  int i, j, k, r, ls, lp, sp, pp;
  while (scanf("%s%s", s, p) == 2) {
    ls = strlen(s);
    lp = strlen(p);
    for (__typeof(ls) i = 0; i < ls; i++) {
      sp = i;
      pp = 0;
      while (sp < ls && pp < lp) {
        if (s[sp] == p[pp]) pp++;
        sp++;
      }
      if (lp == pp)
        A[i] = sp - i;
      else
        A[i] = -1;
    }
    for (__typeof(ls) i = 0; i < ls; i++) {
      for (__typeof(i + 1) r = 0; r < i + 1; r++) {
        if (A[i] >= 0) {
          D[i + A[i]][r + A[i] - lp] =
              max(D[i + A[i]][r + A[i] - lp], D[i][r] + 1);
        }
        D[i + 1][r + 1] = max(D[i + 1][r + 1], D[i][r]);
        D[i + 1][r] = max(D[i + 1][r], D[i][r]);
      }
    }
    printf("%d", D[ls][0]);
    for (__typeof(ls) r = 1; r <= ls; r++) printf(" %d", D[ls][r]);
    printf("\n");
  }
}
