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 dp[2002][2002];
int nc[2002][30];
int mv[2003];
int main() {
  string s, p;
  cin >> s >> p;
  int n = s.size();
  int k = p.size();
  dp[0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int c = 0; c < 30; c++) {
      nc[i][c] = -1;
      for (int ii = i; ii < n; ii++) {
        if (s[ii] - 'a' == c) {
          nc[i][c] = ii;
          break;
        }
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int ii = 0; ii <= n; ii++) {
      if (dp[i][ii] > 0) {
        mv[dp[i][ii]] = max(mv[dp[i][ii]], ii);
        if (i < n) {
          int nci = nc[i][p[ii % k] - 'a'];
          if (nci > -1) {
            if (dp[nci + 1][ii + 1] == 0 ||
                dp[nci + 1][ii + 1] > dp[i][ii] + nci - i) {
              dp[nci + 1][ii + 1] = dp[i][ii] + nci - i;
            }
          }
          if (dp[i + 1][ii - ii % k] == 0 ||
              dp[i + 1][ii - ii % k] > dp[i][ii]) {
            dp[i + 1][ii - ii % k] = dp[i][ii];
          }
        }
      }
    }
  }
  int mmv = 0;
  for (int i = 1; i <= n; i++) {
    mmv = max(mv[i], mmv);
    cout << min(mmv / k, (n - i + 1) / k) << " ";
  }
  cout << 0 << endl;
}
