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>
#pragma warning(disable : 4244 4267 4018 4996 4800)
using namespace std;
istream& in = cin;
ostream& out = cout;
int main() {
  ios_base::sync_with_stdio(false);
  string s, p;
  in >> s >> p;
  vector<vector<int> > v(s.size(), vector<int>(s.size() + 1));
  vector<vector<int> > w(s.size(), vector<int>(p.size(), -1));
  vector<int> last(p.size(), -1);
  for (int i = 0; i < s.size(); ++i) {
    for (int j = p.size() - 1; j >= 0; --j) {
      if (s[i] == p[j]) {
        if (j == 0) {
          w[i][0] = 0;
          last[0] = i;
        } else if (last[j - 1] != -1 && w[last[j - 1]][j - 1] != -1) {
          w[i][j] = i - last[j - 1] - 1 + w[last[j - 1]][j - 1];
          last[j] = i;
        }
      }
    }
  }
  for (int i = 0; i < s.size(); ++i) {
    for (int j = 0; j <= s.size(); ++j) {
      if (i) {
        v[i][j] = v[i - 1][j];
        if (j) v[i][j] = max(v[i][j], v[i - 1][j - 1]);
      }
      int del = w[i][p.size() - 1];
      if (del != -1 && j >= del) {
        int prev = i - del - p.size();
        if (prev == -1 && j - del == 0)
          v[i][j] = 1;
        else if (prev >= 0 && j - del <= prev + 1)
          v[i][j] = max(v[i][j], v[prev][j - del] + 1);
      }
    }
  }
  for (int i = 0; i <= s.size(); ++i) out << v[s.size() - 1][i] << ' ';
  return 0;
}
