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 INF = 1 << 28;
int main() {
  string s, t;
  cin >> s >> t;
  const int n = s.size();
  const int m = t.size();
  int p = 0;
  vector<int> cur(n + 1, INF), nxt(n + 1, INF);
  cur[0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      nxt[j] = min(nxt[j], cur[j] + 1);
      if (j < n && s[i] == t[j % m]) {
        nxt[j + 1] = min(nxt[j + 1], cur[j]);
      }
      nxt[j / m * m] = min(nxt[j / m * m], cur[j]);
    }
    cur.swap(nxt);
    fill(nxt.begin(), nxt.end(), INF);
  }
  for (int x = 0; x <= n; x++) {
    int res = 0;
    for (int i = 0; i <= n; i++) {
      if (cur[i] <= x) {
        res =
            max(res, max(0, (i - max(0, (x - cur[i]) - (n - i - cur[i]))) / m));
      }
    }
    cout << res << " \n"[x == n];
  }
}
