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 f[2005][2005], pre[2005];
int ns, np;
string s, p;
int main() {
  ios::sync_with_stdio(0);
  cin >> s >> p;
  ns = int(s.size()), np = int(p.size());
  for (int i = np - 1; i < ns; i++) {
    int cost = 0, match = 0;
    for (int j = i; j >= 0 && match < np; j--) {
      if (s[j] == p[np - match - 1])
        match++;
      else
        cost++;
    }
    if (match == np) pre[i + 1] = i - match - cost + 2;
  }
  memset(f, -1, sizeof f);
  f[0][0] = 0;
  for (int i = 1; i <= ns; i++)
    for (int j = 0; j <= i; j++) {
      if (i == 4 && j == 0) {
        int t = 0;
      }
      f[i][j] = max(f[i - 1][j - 1], f[i - 1][j]);
      if (pre[i] >= 1) {
        int cost = i - pre[i] + 1 - np;
        if (j >= cost && f[pre[i] - 1][j - cost] != -1)
          f[i][j] = max(f[i][j], f[pre[i] - 1][j - cost] + 1);
      }
    }
  for (int i = 0; i <= ns; i++)
    cout << ((f[ns][i] == -1) ? 0 : f[ns][i]) << ' ';
  cout << '\n';
  return 0;
}
