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;
string s, p;
int n, m, cur, f[2222][2222], b[2222];
int main() {
  ios::sync_with_stdio(false);
  getline(cin, s);
  s = ' ' + s;
  m = int(s.size()) - 1;
  getline(cin, p);
  p = '1' + p;
  n = int(p.size()) - 1;
  for (int i = n; i <= m; ++i) {
    int cur = n + 1;
    for (int j = i; j >= 1; --j)
      if (s[j] == p[cur - 1]) {
        cur--;
        if (cur == 1) b[i] = j;
      }
  }
  for (int i = 1; i <= m; ++i) f[0][i] = -1;
  f[0][0] = 0;
  for (int i = 1; i <= m; ++i)
    for (int j = 0; j <= m - 1; ++j) {
      f[i][j] = max(f[i - 1][j], f[i - 1][j - 1]);
      if (b[i] != 0 && j + n - (i - b[i] + 1) >= 0 &&
          f[b[i] - 1][j + n - (i - b[i] + 1)] != -1)
        f[i][j] = max(f[i][j], f[b[i] - 1][j + n - (i - b[i] + 1)] + 1);
    }
  for (int i = 0; i <= m; ++i) cout << f[m][i] << ' ';
  return 0;
}
