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;
const int N = 2100;
int n, m, dp[N][N], ans[N];
char s[N], p[N];
signed main() {
  scanf("%s%s", s + 1, p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      int k = j % m;
      if (j % m == 0) k = m;
      dp[i][j] = min(dp[i][j], dp[i - 1][j] + (k != m));
      if (s[i] == p[k] && j) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
    }
  }
  for (int i = m; i <= n; i++) {
    for (int j = dp[n][i]; j <= n - (i / m) * m; j++)
      ans[j] = max(ans[j], i / m);
  }
  for (int i = 0; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
}
