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 s1[2200], s2[1000];
int dp[2200][2200];
int cnt[2200];
int main() {
  scanf("%s%s", s1 + 1, s2);
  int len1, len2;
  len1 = strlen(s1 + 1);
  len2 = strlen(s2);
  for (int i = 1; i <= len1; i++) {
    int t1 = i, t2 = len2 - 1;
    while (t1 >= 1 && t2 >= 0) {
      if (s1[t1--] == s2[t2]) {
        t2--;
      }
    }
    if (t2 == -1) {
      cnt[i] = i - t1;
    } else {
      cnt[i] = -1;
    }
  }
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= len1; i++) {
    for (int j = 0; j <= len1; j++) {
      dp[i][j] = max(dp[i - 1][j], dp[i][j]);
      if (cnt[i] != -1 && cnt[i] - len2 <= j &&
          i - cnt[i] >= j + len2 - cnt[i]) {
        dp[i][j] = max(dp[i][j], dp[i - cnt[i]][j + len2 - cnt[i]] + 1);
      }
    }
  }
  for (int i = 0; i <= len1; i++) {
    printf("%d ", dp[len1][i]);
  }
  printf("\n");
  return 0;
}
