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 s[2010], p[510];
int l1, l2, a[2010], dp[2010][2010];
int main() {
  memset(a, 255, sizeof(a));
  scanf("%s%s", s + 1, p + 1);
  l1 = strlen(s + 1);
  l2 = strlen(p + 1);
  for (int i = 1; i <= l1; i++) {
    int np = 1;
    for (int j = i; j <= l1; j++) {
      if (s[j] == p[np]) np++;
      if (np == l2 + 1) {
        a[i] = j - i + 1;
        break;
      }
    }
  }
  for (int i = 0; i <= l1; i++) fill(dp[i], dp[i] + l1 + 1, -1e9);
  dp[0][0] = 0;
  for (int i = 0; i < l1; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (a[i + 1] != -1)
        dp[i + a[i + 1]][j + a[i + 1] - l2] =
            max(dp[i + a[i + 1]][j + a[i + 1] - l2], dp[i][j] + 1);
    }
  }
  for (int i = 0; i <= l1; i++) printf("%d ", dp[l1][i]);
  return 0;
}
