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;
int dp[2003][2003];
int lst[2003];
char s[2003], p[2003];
int n, m;
void calc_last(int right) {
  lst[right] = -1;
  int i = right, j = m;
  while (j >= 1 && i >= 1) {
    if (s[i] == p[j]) {
      j--;
    }
    i--;
  }
  if (j == 0) {
    lst[right] = i + 1;
  }
}
int main() {
  int i, j;
  scanf("%s", s + 1);
  scanf("%s", p + 1);
  n = strlen(s + 1), m = strlen(p + 1);
  for (i = 1; i <= n; ++i) {
    calc_last(i);
  }
  for (i = 1; i <= n; ++i) dp[0][i] = -1000000;
  for (i = 1; i <= n; ++i) {
    for (j = 0; j <= n; ++j) {
      dp[i][j] = -1000000;
      int len = i - lst[i] + 1;
      if (lst[i] != -1 && j >= len - m) {
        dp[i][j] = dp[lst[i] - 1][j - (len - m)] + 1;
      }
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (j > 0) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      }
    }
  }
  for (i = 0; i <= n; ++i) {
    printf("%d ", dp[n][i]);
  }
  return 0;
}
