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;
void solve() {
  string s;
  string p;
  cin >> s >> p;
  int n = s.size();
  reverse(begin(p), end(p));
  vector<vector<int>> dp(n + 1, vector<int>(n + 1, INT_MIN));
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i) {
    dp[i + 1][0] = dp[i][0];
    for (int j = 1; j < n + 1; ++j) {
      dp[i + 1][j] = max(dp[i][j], dp[i][j - 1]);
    }
    if (p[0] == s[i]) {
      int k = 1;
      int ind = i - 1;
      while (k < p.size() && ind >= 0) {
        if (p[k] == s[ind]) ++k;
        --ind;
      }
      ++ind;
      if (k < p.size()) {
        continue;
      }
      int rem = i - ind + 1 - p.size();
      for (int j = rem; j < n + 1; ++j) {
        dp[i + 1][j] = max(dp[i + 1][j], dp[ind][j - rem] + 1);
      }
    }
  }
  for (auto a : dp.back()) {
    cout << a << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();
  return 0;
}
