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 go[4025], f[4025][4025], n, m;
char s1[4025], s2[4025];
void init() {
  scanf("%s", s1 + 1);
  n = strlen(s1 + 1);
  scanf("%s", s2 + 1);
  m = strlen(s2 + 1);
  for (int i = m; i <= n; i++) {
    int now = i, flag = 1;
    for (int j = m; j >= 1; j--) {
      while (now && s1[now] != s2[j]) now--;
      if (!now) {
        flag = 0;
        break;
      }
      now--;
    }
    if (flag) go[i] = now + 1;
  }
}
void work() {
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) f[i][j] = -(1 << 29);
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= i; j++)
      if (go[i]) {
        int x = j - (i - go[i] + 1 - m);
        if (x >= 0) f[i][j] = max(f[i][j], f[go[i] - 1][x] + 1);
      }
    for (int j = 0; j <= i; j++) {
      if (j) f[i][j] = max(f[i][j], f[i - 1][j - 1]);
      f[i][j] = max(f[i][j], f[i - 1][j]);
    }
  }
  for (int i = 0; i <= n; i++) printf("%d ", f[n][i]);
}
int main() {
  init();
  work();
}
