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 s[2005], t[2005];
int dp[2005][2005], ans[2005];
int main() {
  int T, i, j, ca = 0, m, n, k;
  while (~scanf("%s%s", s, t)) {
    n = strlen(s), m = strlen(t);
    k = n / m;
    for (i = 0; i <= n; i++)
      for (j = 0; j <= n; j++) dp[i][j] = 1000000007;
    for (i = 0; i <= n; i++) dp[i][0] = 0;
    for (i = 0; i < n; i++)
      for (j = 0; j < k * m; j++) {
        if (s[i] == t[j % m]) {
          dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j]);
        }
        dp[i + 1][j + 1] =
            min(dp[i + 1][j + 1], dp[i][j + 1] + (j % m == m - 1 ? 0 : 1));
      }
    for (i = 0; i <= n; i++) ans[i] = 0;
    for (j = k; j >= 0; j--) {
      int first = dp[n][j * m], second = n - j * m;
      for (i = first; i <= second; i++) ans[i] = max(ans[i], j);
    }
    for (i = 0; i <= n; i++) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}
