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 main() {
  ios_base::sync_with_stdio(0);
  string s, wzo;
  cin >> s >> wzo;
  int n = s.size();
  int m = wzo.size();
  vector<int> pod(n);
  for (int i = 0; i < n; ++i) {
    int j = i, k = m - 1;
    while (j >= 0 && k >= 0) {
      if (s[j] == wzo[k]) --k;
      --j;
    }
    if (k == -1)
      pod[i] = j + 1;
    else
      pod[i] = -1;
  }
  vector<vector<int> > dp(n, vector<int>(n + 1, 0));
  if (m == 1 && wzo[0] == s[0]) dp[0][0] = 1;
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j <= i + 1; ++j) {
      if (j <= i) dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (j != 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      if (pod[i] != -1) {
        int usu = i - pod[i] + 1 - m;
        if (usu <= j) {
          int resz = j - usu;
          if (resz <= pod[i]) {
            int wcz = 0;
            if (pod[i] != 0) wcz = dp[pod[i] - 1][resz];
            dp[i][j] = max(dp[i][j], wcz + 1);
          }
        }
      }
    }
  }
  for (int i = 0; i <= n; ++i) cout << dp[n - 1][i] << " ";
  return 0;
}
