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 N = 2008;
const int M = 1000000008;
char a[N], b[N];
int n, m, mm, c[N], d[N], f[N][N], ans[N];
int minx(int x, int y) {
  if (x < y) y = x;
  return y;
}
int main(void) {
  int i, j, k;
  scanf("%s%s", a + 1, b + 1);
  n = strlen(a + 1);
  m = strlen(b + 1);
  for (i = 1; i <= n; i++) {
    c[i] = 0;
    k = 1;
    for (j = i; j <= n && k <= m; j++)
      if (a[j] == b[k]) {
        k++;
      } else {
        c[i]++;
      }
    if (k <= m) {
      c[i] = d[i] = M;
    } else {
      d[i] = j - 1;
    }
  }
  mm = n / m;
  for (i = 1; i <= n + 1; i++) {
    f[i][0] = 0;
    for (j = 1; j <= mm; j++) f[i][j] = M;
  }
  for (i = n; i >= 1; i--) {
    for (j = 1; j <= mm; j++) {
      f[i][j] = f[i + 1][j];
      if (c[i] < M && i + m - 1 <= n)
        f[i][j] = minx(f[i][j], c[i] + f[d[i] + 1][j - 1]);
    }
  }
  memset(ans, -1, sizeof(ans));
  ans[0] = 0;
  for (i = 0; i <= mm; i++)
    if (f[1][i] < M) ans[f[1][i]] = i;
  for (i = 1; i <= n; i++)
    if (ans[i] == -1) ans[i] = minx(ans[i - 1], (n - i) / m);
  for (i = 0; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}
