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;
const int INF = 1e9;
int main() {
  ios_base::sync_with_stdio(0);
  string s, t;
  cin >> s >> t;
  int n = s.size();
  int m = t.size();
  vector<vector<int> > dp(n + 1, vector<int>(n + 2, -INF));
  dp[0][0] = 0;
  vector<int> ans(n + 1);
  vector<int> p(m + 1);
  for (int i = 1; i < m; i++) {
    int j = p[i - 1];
    while (j && t[j] != t[i]) {
      j = p[j - 1];
    }
    if (t[j] == t[i]) {
      j++;
    }
    p[i] = j;
  }
  vector<vector<int> > to(m + 1, vector<int>('z' + 1));
  for (int i = 0; i < m; i++) {
    for (int c = 'a'; c <= 'z'; c++) {
      to[i][c] = to[p[max(0, i - 1)]][c];
    }
    to[i][t[i]] = i + 1;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      int j2 = j - j % m + to[j % m][s[i]];
      dp[i + 1][j2] = max(dp[i][j] + 1, dp[i + 1][j2]);
      dp[i + 1][j] = max(dp[i][j], dp[i + 1][j]);
    }
  }
  for (int j = 0; j <= n; j++) {
    for (int i = (j / m) * m; i <= dp[n][j]; i++) {
      ans[i] = max(ans[i], j / m);
    }
  }
  for (int i = n; i >= 0; i--) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}
