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 = 2e3 + 100;
int dp[maxn][maxn];
char s[maxn], p[maxn];
int s_len, p_len;
void solve() {
  memset(dp, 0, sizeof(dp));
  s_len = strlen(s + 1);
  p_len = strlen(p + 1);
  for (int i = 1; i <= s_len; i++) {
    int s_p = i, p_p = p_len;
    int cnt = 0;
    while (p_p != 0 && s_p != 0) {
      if (s[s_p] == p[p_p]) {
        s_p--;
        p_p--;
      } else {
        cnt++;
        s_p--;
      }
    }
    for (int j = 0; j <= i; j++) {
      if (p_p == 0 && j >= cnt) {
        if (s_p >= j - cnt) dp[i][j] = max(dp[i][j], 1 + dp[s_p][j - cnt]);
      }
      if (i - 1 >= j) dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (j >= 1) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
    }
  }
  for (int i = 0; i <= s_len; i++) {
    printf("%d%c", dp[s_len][i], (i == s_len ? '\n' : ' '));
  }
}
int main() {
  while (scanf("%s%s", s + 1, p + 1) != EOF) {
    solve();
  }
  return 0;
}
