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 maxn = 2000 + 5;
char s[maxn], p[maxn];
int lens, lenp, pos;
int dp[maxn][maxn];
int cal(int ps) {
  int lp = lenp;
  int cnt = 0;
  for (int i = ps; i >= 1; i--) {
    if (s[i] == p[lp]) {
      lp--;
      if (lp == 0) {
        pos = i;
        return cnt;
      }
    } else
      cnt++;
  }
}
int main() {
  scanf("%s", s + 1);
  lens = strlen(s + 1);
  scanf("%s", p + 1);
  lenp = strlen(p + 1);
  for (int i = lenp; i <= lens; i++) {
    int cnt = cal(i);
    for (int j = 0; j < i; j++) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (j >= cnt && pos - 1 >= j - cnt) {
        dp[i][j] = max(dp[i][j], dp[pos - 1][j - cnt] + 1);
      }
    }
  }
  for (int i = 0; i <= lens; i++) {
    printf("%d%c", dp[lens][i], i == lens ? '\n' : ' ');
  }
  return 0;
}
