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 = 2005, mod = 1e9 + 7;
int dp[N][N];
char s[N], p[N];
int main() {
  int i, n, m, j, k, T, l, r;
  int ans, t, x, y;
  while (~scanf("%s%s", s + 1, p + 1)) {
    memset(dp, 0, sizeof dp);
    n = strlen(s + 1), m = strlen(p + 1);
    for (i = 1; i <= n; i++) {
      for (x = i, k = m; x && k; x--)
        if (s[x] == p[k]) k--;
      for (x++, j = 0; j <= i - m; j++)
        if (!k && j - i + x + m - 1 >= 0)
          dp[i][j] = dp[x - 1][j - i + x + m - 1] + 1;
      for (j = 0; j <= i; j++) {
        if (j) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
        dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      }
    }
    for (i = 0; i <= n; i++) printf("%d ", dp[n][i]);
    printf("\n");
  }
  return 0;
}
