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 n, p, g[2010UL][2010UL], f[2010UL][2010UL];
char s1[2010UL], s2[2010UL];
int main() {
  scanf("%s%s", s1 + 1, s2 + 1);
  n = strlen(s1 + 1), p = strlen(s2 + 1);
  memset(g, -1, sizeof(g));
  for (int i = 0; i < n; ++i) {
    g[i][0] = i;
    for (int j = 0; j < p; ++j)
      if (g[i][j] != -1) {
        g[i + 1][j] = max(g[i + 1][j], g[i][j]);
        if (s1[i + 1] == s2[j + 1])
          g[i + 1][j + 1] = max(g[i + 1][j + 1], g[i][j]);
      }
  }
  memset(f, -0x3f, sizeof(f));
  f[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= i; ++j) {
      f[i][j] = f[i - 1][j];
      if (j > 0) f[i][j] = max(f[i][j], f[i - 1][j - 1]);
      if (g[i][p] != -1) {
        int nw = i - g[i][p] - p;
        if (j - nw >= 0) f[i][j] = max(f[i][j], f[g[i][p]][j - nw] + 1);
      }
    }
  }
  for (int i = 0; i <= n; ++i) printf("%d ", f[n][i]);
  return 0;
}
