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 str1[2011], str2[2011];
int dp[2011][2011];
int main() {
  scanf("%s%s", str1 + 1, str2 + 1);
  int len1 = strlen(str1 + 1);
  int len2 = strlen(str2 + 1);
  for (int i = 1; i <= len1; i++) {
    int ii = i, jj = len2;
    while (ii >= 1 && jj >= 1) {
      if (str1[ii] == str2[jj]) jj--;
      ii--;
    }
    if (jj == 0) {
      int dif = i - ii - len2;
      for (int th = 0; th <= ii; th++) {
        dp[i][dif + th] = dp[ii][th] + 1;
      }
    }
    for (int th = 0; th < i; th++) {
      dp[i][th] = max(dp[i][th], dp[i - 1][th]);
    }
  }
  for (int i = 0; i <= len1; i++) {
    printf("%d ", dp[len1][i]);
  }
  return 0;
}
