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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return l + rng() % (r - l + 1); }
const int N = 5555;
int n, m;
int f[N][N];
string a, b;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  ;
  cin >> a;
  n = a.size();
  a = " " + a;
  cin >> b;
  m = b.size();
  b = " " + b;
  for (int i = 1, _c = n; i <= _c; ++i) f[0][i] = -n - n;
  for (int i = 0, _c = n - 1; i <= _c; ++i) {
    int j = i + 1, t = m;
    while (j && t) {
      if (a[j] == b[t]) t--;
      j--;
    }
    if (!t) {
      for (int k = 0, _c = i + 1; k <= _c; ++k) {
        if (j >= k - i + j - 1 + m && k - i + j - 1 + m >= 0) {
          f[i + 1][k] = max(f[i + 1][k], f[j][k - i + j - 1 + m] + 1);
        }
      }
    }
    for (int k = 0, _c = i + 1; k <= _c; ++k)
      f[i + 1][k] = max(f[i + 1][k], f[i][k]);
  }
  for (int i = 0, _c = n; i <= _c; ++i) {
    cout << f[n][i] << " ";
  }
}
