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 pd[2010][2010];
int pos[2010];
int cont[2010];
char texto[2010];
char padrao[510];
int n, m;
int busca(int lugar, int quant) {
  if (pd[lugar][quant] != -1) {
    return pd[lugar][quant];
  }
  if (lugar == 0) {
    if (quant == 0) {
      return 0;
    } else {
      return -99999999;
    }
  }
  pd[lugar][quant] = -99999999;
  if (pos[lugar] != -1 && cont[lugar] <= quant) {
    pd[lugar][quant] =
        max(pd[lugar][quant], busca(pos[lugar] - 1, quant - cont[lugar]) + 1);
  }
  if (quant > 0) {
    pd[lugar][quant] = max(pd[lugar][quant], busca(lugar - 1, quant - 1));
  }
  pd[lugar][quant] = max(pd[lugar][quant], busca(lugar - 1, quant));
  return pd[lugar][quant];
}
int main() {
  scanf("%s", texto + 1);
  scanf("%s", padrao + 1);
  n = strlen(texto + 1);
  m = strlen(padrao + 1);
  int i, j, k;
  for (i = 1; i <= n; i++) {
    j = i;
    k = m;
    while (1) {
      if (j == 0 || k == 0) {
        break;
      }
      if (texto[j] == padrao[k]) {
        j--;
        k--;
      } else {
        j--;
      }
    }
    if (k == 0) {
      pos[i] = j + 1;
      cont[i] = i - pos[i] + 1 - m;
    } else {
      pos[i] = -1;
    }
  }
  memset(pd, -1, sizeof(pd));
  for (i = 0; i <= n; i++) {
    if (i != 0) {
      printf(" ");
    }
    printf("%d", busca(n, i));
  }
  printf("\n");
  return 0;
}
