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;
const int nmax = 2000 + 18, mmax = 500 + 18;
int n, m;
char s[nmax], p[mmax];
int g[nmax], f[nmax][nmax];
inline void update(int &a, int b) {
  if (a < b) a = b;
}
int main() {
  memset(f, -1, sizeof(f));
  scanf("%s%s", s + 1, p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  for (int i = 1; i <= n; ++i) {
    int j = 1;
    g[i] = -1;
    for (int k = i; k <= n; ++k) {
      if (s[k] == p[j]) ++j;
      if (j > m) {
        g[i] = k;
        break;
      }
    }
  }
  f[0][0] = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= n; ++j)
      if (f[i][j] != -1) {
        update(f[i + 1][j], f[i][j]);
        if (j + 1 <= n) update(f[i + 1][j + 1], f[i][j]);
        if (g[i + 1] != -1 && j + g[i + 1] - i - m <= n)
          update(f[g[i + 1]][j + g[i + 1] - i - m], f[i][j] + 1);
      }
  for (int i = 0; i <= n; ++i) printf("%d%c", f[n][i], i == n ? '\n' : ' ');
  return 0;
}
