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>
char s[2010], p[510];
int dp[2010][2010];
int pt[2010][510];
int latest[510];
int ans[2010];
int max(int a, int b) { return a > b ? a : b; }
int main() {
  int n, m, i, j;
  scanf("%s%s", s, p);
  n = strlen(s);
  m = strlen(p);
  for (j = 0; j < m; j++) latest[j] = -1;
  for (i = 0; i < n; i++) {
    for (j = m - 1; j >= 0; j--) {
      pt[i][j] = -1;
      if (s[i] == p[j]) {
        if (j == 0) {
          pt[i][j] = i;
          latest[0] = i;
        } else {
          if (latest[j - 1] != -1) {
            pt[i][j] = pt[latest[j - 1]][j - 1];
            latest[j] = i;
          }
        }
      }
    }
  }
  dp[0][0] = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= i; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j != 0) dp[i][j] = max(dp[i - 1][j - 1], dp[i][j]);
      if (pt[i - 1][m - 1] == -1) continue;
      int p = j - (i - pt[i - 1][m - 1] - m);
      if (p >= 0 && pt[i - 1][m - 1] >= p)
        dp[i][j] = max(dp[i][j], dp[pt[i - 1][m - 1]][p] + 1);
    }
  }
  for (i = 0; i <= n; i++) printf("%d ", dp[n][i]);
  return 0;
}
