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>
#pragma warning(disable : 4996)
using namespace std;
char a[10000], b[10000];
int n, m;
int d[2005][2005];
int calc(int now) {
  int i, j;
  int nowb = m - 1;
  for (i = now; i >= 0; i--) {
    if (a[i] == b[nowb]) {
      nowb--;
      if (nowb < 0) break;
    }
  }
  if (nowb >= 0) return -1;
  return i;
}
int main() {
  int i, j, k;
  scanf("%s%s", a, b);
  n = strlen(a), m = strlen(b);
  for (i = 0; i < n; i++) {
    int last = calc(i);
    if (last != -1) {
      if (last == 0) {
        d[i][i - last + 1 - m] = 1;
      } else {
        for (j = 0; j <= last; j++) {
          d[i][j + i - last + 1 - m] = d[last - 1][j] + 1;
        }
      }
    }
    for (j = 0; j <= n; j++) {
      if (i) d[i][j] = max(d[i][j], d[i - 1][j]);
      if (i && j) d[i][j] = max(d[i][j], d[i - 1][j - 1]);
    }
  }
  for (i = 0; i <= n; i++) printf("%d ", d[n - 1][i]);
  return 0;
}
