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[2010][2010];
int main() {
  string a, b;
  cin >> a >> b;
  int i, k, j, del, s = a.size(), p = b.size();
  for (i = 1; i <= s; i++) {
    k = p;
    for (j = i; j && k; j--)
      if (a[j - 1] == b[k - 1]) k--;
    if (k == 0) {
      del = i - j - p;
      for (k = del; k <= i - p; k++) f[i][k] = max(f[i][k], f[j][k - del] + 1);
    }
    for (k = 0; k < i; k++) f[i][k] = max(f[i][k], f[i - 1][k]);
  }
  for (i = 0; i < s; i++) printf("%d ", f[s][i]);
  puts("0");
}
