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[2005], s2[2005];
int l1, l2, j, d[2005][2005], f;
int mac(int i) {
  int x = i, y = l2;
  while (x && y) {
    if (!y) break;
    if (s1[x - 1] == s2[y - 1]) {
      x--;
      y--;
    } else
      x--;
  }
  if (!y) {
    j = x;
    return 1;
  } else {
    j = 0;
    return 0;
  }
}
int main() {
  scanf("%s", s1);
  scanf("%s", s2);
  l1 = strlen(s1);
  l2 = strlen(s2);
  for (int i = 1; i <= l1; i++) {
    f = mac(i);
    for (int k = 0; k <= i; k++) {
      d[i][k] = max(d[i - 1][k], d[i - 1][k - 1]);
      if (f && j >= k - i + j + l2 && k - i + j + l2 >= 0)
        d[i][k] = max(d[i][k], d[j][k - (i - j - l2)] + 1);
    }
  }
  for (int i = 0; i <= l1; i++) printf("%d ", d[l1][i]);
  return 0;
}
