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;
int nxt[2010], dp[2010][2010];
int main() {
  int n, m, i, j;
  string s, p;
  cin >> s >> p;
  n = s.size(), m = p.size();
  s = " " + s;
  for (i = 0; i < n; i++) {
    int cur = 0;
    for (j = i + 1; j <= n; j++) {
      if (s[j] == p[cur]) cur++;
      if (cur == m) break;
    }
    nxt[i] = j;
  }
  memset(dp, 0xc0, sizeof(dp));
  dp[0][0] = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j <= n; j++) {
      if (j != n) dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (nxt[i] != n + 1)
        dp[nxt[i]][j + m] = max(dp[nxt[i]][j + m], dp[i][j] + 1);
    }
  }
  for (i = n; i >= 0; i--) printf("%d ", dp[n][i]);
}
