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 MAXN = 2222;
char s[MAXN], p[MAXN];
int last[MAXN];
int f[MAXN][MAXN];
void Init() {
  assert(scanf(" %s", s + 1) == 1);
  assert(scanf(" %s", p + 1) == 1);
}
void Work() {
  int n, m;
  n = strlen(s + 1);
  m = strlen(p + 1);
  for (int i = 1; i <= n; ++i) {
    last[i] = -1;
    for (int j = i, t = m; j >= 0; --j) {
      if (!t) {
        last[i] = j;
        break;
      }
      if (s[j] == p[t]) --t;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= i; ++j) {
      f[i][j] = f[i - 1][j];
      if (last[i] != -1) {
        int k = j - (i - last[i] - m);
        if (0 <= k && k <= last[i])
          if (f[i][j] < f[last[i]][k] + 1) f[i][j] = f[last[i]][k] + 1;
      }
    }
  }
  for (int i = 0; i <= n; ++i) {
    printf("%d ", f[n][i]);
  }
  printf("\n");
}
int main() {
  Init();
  Work();
  return 0;
}
