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 f[2001][2001], n, m;
char str[2222], str1[2222];
int main() {
  memset(f, 0, sizeof(f));
  scanf("%s%s", str + 1, str1 + 1);
  n = strlen(str + 1);
  m = strlen(str1 + 1);
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; i++) {
    int Left = 1, k = i;
    for (; Left <= m && k <= n; k++)
      if (str[k] == str1[Left])
        if (++Left == m + 1) break;
    for (int j = 0; j < i; j++) {
      if (k <= n)
        f[k][j + k - i + 1 - m] = max(f[k][j + k - i + 1 - m], f[i - 1][j] + 1);
      else
        f[n][j + n - i + 1] = max(f[n][j + n - i + 1], f[i - 1][j]);
      f[i][j] = max(f[i][j], f[i - 1][j]);
    }
  }
  for (int i = 0; i <= n; i++) {
    printf("%d", f[n][i]);
    if (i != n) printf(" ");
  }
  printf("\n");
}
