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;
vector<vector<int> > v;
vector<int> pot;
int n, m;
int fun(int pos, int oc) {
  if (v[pos][oc] != -1) return v[pos][oc];
  if (pos == n) {
    if (oc == 0)
      return v[pos][oc] = 0;
    else
      return v[pos][oc] = -2000000;
  }
  v[pos][oc] = fun(pos + 1, oc);
  if (oc) v[pos][oc] = max(v[pos][oc], fun(pos + 1, oc - 1));
  if (pot[pos] != -1 && oc >= pot[pos] - m - pos)
    v[pos][oc] = max(v[pos][oc], fun(pot[pos], oc - (pot[pos] - pos - m)) + 1);
  return v[pos][oc];
}
int main() {
  ios_base::sync_with_stdio(false);
  string s, p;
  cin >> s >> p;
  n = s.size();
  m = p.size();
  v = vector<vector<int> >(n + 1, vector<int>(n + 1, -1));
  pot = vector<int>(n + 1, -1);
  for (int i = 0; i < s.size(); i++) {
    if (i + p.size() > s.size())
      pot[i] = -1;
    else {
      int ii = i;
      for (int j = 0; j < p.size(); j++) {
        while (ii < s.size() and s[ii] != p[j]) ++ii;
        if (j + 1 == p.size() and ii < s.size()) pot[i] = ii + 1;
        ++ii;
      }
    }
  }
  for (int i = 0; i < s.size() + 1; i++) {
    if (i) cout << ' ';
    cout << fun(0, i);
  }
  cout << '\n';
}
