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 a[3000], b[1000];
int f[3000][3000];
int main() {
  int n, m;
  scanf("%s%s", a + 1, b + 1);
  n = strlen(a + 1);
  m = strlen(b + 1);
  for (int i = 1; i <= n; ++i) {
    int k = m;
    int j;
    for (j = i; j > 0 && k; --j)
      if (a[j] == b[k]) k--;
    if (k == 0)
      for (k = 0; k <= i; ++k)
        if (k + j + m >= i) f[i][k] += f[j][k + j + m - i] + 1;
    for (k = 0; k <= i; ++k) {
      if (f[i][k] < f[i - 1][k]) f[i][k] = f[i - 1][k];
      if (f[i][k] < f[i][k - 1]) f[i][k] = f[i][k - 1];
    }
  }
  for (int i = 0; i <= n; ++i) printf("%d ", min(f[n][i], (n - i) / m));
}
