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;
signed main() {
  ios::sync_with_stdio(false);
  string S, P;
  cin >> S >> P;
  vector<int> word(S.size());
  for (int i = 0; i < S.size(); ++i) {
    int j = i, p = 0;
    for (; p < P.size() && j < S.size(); ++j) {
      if (P[p] == S[j]) {
        ++p;
      }
    }
    if (p == P.size()) {
      word[i] = j - i;
    } else {
      word[i] = -1;
    }
  }
  vector<vector<int>> dp(S.size() + 1, vector<int>(S.size() + 1));
  for (int i = 0; i < S.size(); ++i) {
    for (int j = 0; j <= i; ++j) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (j + 1 <= S.size()) {
        dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      }
      if (~word[i] && i + word[i] <= S.size() &&
          j + word[i] - P.size() <= S.size()) {
        dp[i + word[i]][j + word[i] - P.size()] =
            max(dp[i + word[i]][j + word[i] - P.size()], dp[i][j] + 1);
      }
    }
  }
  for (int i = 0; i <= S.size(); ++i) {
    cout << dp[S.size()][i] << " \n"[i == S.size()];
  }
  return 0;
}
