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 Dp[3001][3001];
int main(void) {
  ios::sync_with_stdio(false);
  int i, j, k;
  string S1, S2;
  cin >> S1 >> S2;
  int L1, L2;
  L1 = S1.size(), L2 = S2.size();
  for (i = 1; i <= L1; i++) {
    k = L2;
    for (j = i; j && k; j--) {
      if (S1[j - 1] == S2[k - 1]) {
        k--;
      }
    }
    if (!k) {
      int Delta;
      Delta = i - L2 - j;
      for (k = Delta; k + L2 <= i; k++)
        Dp[i][k] = max(Dp[i][k], Dp[j][k - Delta] + 1);
    }
    for (k = 0; k < i; k++) Dp[i][k] = max(Dp[i][k], Dp[i - 1][k]);
  }
  for (i = 0; i <= L1; i++) {
    cout << Dp[L1][i] << ' ';
  }
  cout << endl;
  return 0;
}
