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 MAX = 2000 + 10;
char s[MAX], p[MAX];
int dp[MAX][MAX];
int main() {
  int n, m;
  while (scanf("%s %s", s + 1, p + 1) == 2) {
    n = strlen(s + 1);
    m = strlen(p + 1);
    memset(dp, -1, sizeof(dp));
    dp[0][0] = 0;
    for (int i = 1; i <= n; i++) {
      int first = i, last = m;
      while (first > 0 && last > 0) {
        if (p[last] == s[first]) {
          last--;
        }
        first--;
      }
      for (int j = 0; j <= i; j++) {
        if (dp[i - 1][j] != -1) dp[i][j] = dp[i - 1][j];
        if (j > 0 && dp[i - 1][j - 1] != -1)
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
        if (last == 0 && i - first - m <= j &&
            dp[first][j - (i - first - m)] != -1) {
          dp[i][j] = max(dp[i][j], dp[first][j - (i - first - m)] + 1);
        }
      }
    }
    for (int i = 0; i <= n; i++) {
      printf("%d ", dp[n][i]);
    }
    printf("\n");
  }
}
