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[2100], s2[2100];
int get(int p) {
  int len = strlen(s2) - 1;
  while (len >= 0) {
    while (p >= 0 && s1[p] - s2[len]) p--;
    if (p < 0) return -1;
    len--, p--;
  }
  return p + 1;
}
int dp[2100][2100];
int pre[2100];
int main() {
  int i, j, k;
  scanf("%s", s1);
  scanf("%s", s2);
  for (i = 0; s1[i]; i++) pre[i] = get(i);
  int len2 = strlen(s2);
  for (i = 0; s1[i]; i++) {
    for (j = 0; j <= i + 1; j++) {
      k = i - pre[i] + 1 - len2;
      if (pre[i] >= 0 && j >= k) {
        if (pre[i] > 0 && j - k <= pre[i])
          dp[i][j] = max(dp[i][j], dp[pre[i] - 1][j - k] + 1);
        else if (j == k)
          dp[i][j] = max(dp[i][j], 1);
      }
      if (i) dp[i][j] = max(dp[i][j], dp[i - 1][j]);
    }
  }
  int len = strlen(s1);
  for (i = 0; i <= len; i++) printf("%d ", dp[len - 1][i]);
  puts("");
}
