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 ans[2010][2010];
int main() {
  std::ios_base::sync_with_stdio(false);
  string str, pat;
  int i, j, k, l;
  cin >> str >> pat;
  l = str.length();
  int req[l + 2];
  for (i = 0; i < l + 2; i++) req[i] = -1;
  for (i = 0; i < l; i++) {
    for (j = 0, k = i; j < pat.length() && k < l; j++) {
      while (k < l && str[k] != pat[j]) k++;
      if (k < l)
        k++;
      else
        break;
    }
    if (j == pat.length()) req[i] = (k - i);
  }
  for (i = 0; i < l; i++) {
    for (j = 0; j <= i; j++) {
      ans[i + 1][j] = max(ans[i][j], ans[i + 1][j]);
      ans[i + 1][j + 1] = max(ans[i][j], ans[i + 1][j + 1]);
      if (req[i] >= 0)
        ans[i + req[i]][j + req[i] - pat.length()] =
            max(ans[i + req[i]][j + req[i] - pat.length()], ans[i][j] + 1);
    }
  }
  for (i = 0; i <= l; i++) {
    cout << ans[l][i] << " ";
  }
  return 0;
}
