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 mod = 998244353;
const int maxn = 2000 + 5;
char s[maxn], t[maxn];
int f[maxn][maxn], Next[maxn][26], pre[maxn], pos[26];
int main() {
  scanf("%s%s", s + 1, t + 1);
  int n = strlen(s + 1), m = strlen(t + 1);
  for (int i = n; i >= 0; i--) {
    for (int j = 0; j < 26; j++) Next[i][j] = pos[j];
    if (i) pos[s[i] - 'a'] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (s[i] != t[1]) continue;
    int p = i;
    for (int j = 2; j <= m; j++) {
      p = Next[p][t[j] - 'a'];
      if (!p) break;
    }
    if (!p) continue;
    pre[p] = i;
  }
  memset(f, 0xcf, sizeof(f));
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      f[i][j] = max(f[i - 1][j], f[i - 1][j - 1]);
    }
    if (!pre[i]) continue;
    int tmp = i - pre[i] + 1 - m;
    for (int j = tmp; j <= n; j++) {
      f[i][j] = max(f[i][j], f[pre[i] - 1][j - tmp] + 1);
    }
  }
  for (int i = 0; i <= n; i++) printf("%d ", f[n][i]);
  printf("\n");
  return 0;
}
