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 res[2010][2010], i, j, k, l1, l2, cnt;
int main() {
  string s, p;
  cin >> s >> p;
  l1 = s.length();
  l2 = p.length();
  memset(res, 0, sizeof(res));
  if (l2 == 1 && p[0] == s[0]) res[0][0] = 1;
  for (i = 1; i < l1; ++i) {
    res[i][0] = res[i - 1][0];
    for (j = 1; j <= (i + 1); ++j)
      res[i][j] = max(res[i - 1][j], res[i - 1][j - 1]);
    k = l2 - 1;
    if (s[i] == p[k]) {
      cnt = 0;
      j = i;
      for (; j >= 0; --j) {
        if (s[j] == p[k]) {
          --k;
          if (k < 0) break;
        } else
          ++cnt;
      }
      if (j > 0) {
        for (k = cnt; (k - cnt) <= j; ++k) {
          if (res[j - 1][k - cnt] + 1 > res[i][k]) {
            res[i][k] = res[j - 1][k - cnt] + 1;
          }
        }
      } else if (j == 0) {
        if (res[i][cnt] < 1) res[i][cnt] = 1;
      }
    }
  }
  for (i = 0; i <= l1; ++i) printf("%d ", res[l1 - 1][i]);
}
