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 n, m;
string s, p;
vector<vector<int> > f, g;
int main() {
  cin >> s >> p;
  n = s.size();
  m = p.size();
  f.resize(n + 1, vector<int>(m + 1, INF));
  f[n][m] = n;
  for (int i = n - 1; i >= 0; --i) {
    f[i][m] = i;
    for (int j = 0; j < m; ++j) {
      if (s[i] == p[j])
        f[i][j] = f[i + 1][j + 1];
      else
        f[i][j] = f[i + 1][j];
    }
  }
  g.resize(n + 1, vector<int>(n + 1, -INF));
  g[n][0] = 0;
  for (int i = n - 1; i >= 0; --i) {
    for (int j = 0; j <= n - i; ++j) {
      g[i][j] = max(0, g[i + 1][j]);
      if (f[i][0] - i - m <= j)
        g[i][j] = max(g[i][j], g[f[i][0]][j - (f[i][0] - i - m)] + 1);
    }
  }
  for (int i = 0; i <= n; ++i) printf("%d%c", g[0][i], " \n"[i == n]);
  return 0;
}
