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[2222], s2[2222];
int T[2222];
int f[2222][2222];
int main(int argc, char *argv[]) {
  scanf("%s%s", s1, s2);
  int n = strlen(s1), m = strlen(s2);
  for (int i = 0; i < n; i++)
    if (s1[i] == s2[m - 1]) {
      int j = i, k = m - 1;
      while (k >= 0 && j >= 0) {
        if (s1[j] == s2[k]) {
          j--, k--;
        } else {
          j--;
          T[i]++;
        }
      }
      if (k != -1) {
        T[i] = -1;
      }
    }
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    f[i][0] = f[i - 1][0];
    if (s1[i - 1] == s2[m - 1]) {
      if (T[i - 1] == 0) {
        f[i][0] = max(f[i][0], f[i - m][0] + 1);
      }
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++) {
      if (i - 1 >= j) {
        f[i][j] = max(f[i][j], f[i - 1][j]);
      }
      f[i][j] = max(f[i][j], f[i - 1][j - 1]);
      if (s1[i - 1] == s2[m - 1]) {
        if (i - m - T[i - 1] >= 0 && T[i - 1] >= 0 && T[i - 1] <= j &&
            i - m - T[i - 1] >= j - T[i - 1]) {
          f[i][j] = max(f[i - m - T[i - 1]][j - T[i - 1]] + 1, f[i][j]);
        }
      }
    }
  for (int j = 0; j <= n; j++) {
    printf("%d", f[n][j]);
    printf(j == n ? "\n" : " ");
  }
  return 0;
}
