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 maxl = 2050, maxp = 550;
char s[maxl], p[maxl];
int ls, lp, ans[maxl];
int f[maxl][maxl];
int match(int x) {
  int i, j;
  for (i = x, j = 0; i < ls && j < lp; i++, j++)
    while (i < ls && s[i] != p[j]) i++;
  return j < lp ? 0 : i;
}
int main() {
  int i, j, to;
  scanf("%s%s", s, p);
  ls = strlen(s);
  lp = strlen(p);
  for (i = 0; i < ls; i++) {
    if ((to = match(i))) {
      int t = to - i - lp;
      for (j = 0; j <= i; j++) f[to][j + t] = max(f[to][j + t], f[i][j] + 1);
    }
    for (j = 0; j <= i; j++) f[i + 1][j] = max(f[i + 1][j], f[i][j]);
  }
  ans[0] = f[ls][0];
  for (i = 1; i <= ls && f[ls][i] >= f[ls][i - 1]; i++) ans[i] = f[ls][i];
  for (j = ls - i - ans[i - 1] * lp; i <= ls && j >= 0; i++, j--)
    ans[i] = ans[i - 1];
  for (j = ans[i - 1] * lp - 1; j >= 0; i++, j--) ans[i] = j / lp;
  for (i = 0; i <= ls; i++) printf("%d%c", ans[i], i < ls ? ' ' : '\n');
  return 0;
}
