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(int argc, char* argv[]) {
  cin.sync_with_stdio(false);
  string s;
  string p;
  cin >> s >> p;
  int n = s.size();
  int p_len = p.size();
  vector<int> neededmiss(n + 1, 100000000);
  vector<vector<int>> sol(n + 1, vector<int>(n + 1, 0));
  for (int b = 0; b <= n; ++b) {
    bool ok = true;
    int miss = 0;
    for (int i = p_len - 1; i >= 0; --i) {
      int index = b - 1 - (p_len - 1) + i - miss;
      if (index < 0) {
        ok = false;
        break;
      }
      if (s[index] != p[i]) {
        ++i;
        ++miss;
      }
    }
    if (ok) {
      neededmiss[b] = miss;
    }
  }
  for (int x = 0; x <= n; ++x) {
    for (int b = 0; b <= n; ++b) {
      if (b < p_len + x) {
        sol[x][b] = 0;
      } else {
        sol[x][b] = sol[x][b - 1];
        if (neededmiss[b] <= x) {
          sol[x][b] = max(
              sol[x][b], 1 + sol[x - neededmiss[b]][b - p_len - neededmiss[b]]);
        }
      }
    }
    cout << sol[x][n] << " ";
  }
  cout << endl;
  return 0;
}
