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;
string s, p;
int dp[2010][2010], m, n;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s;
  cin >> p;
  n = s.size();
  m = p.size();
  dp[0][0] = 0;
  s = "Q" + s;
  p = "Q" + p;
  for (int i = 1; i <= n; i++) dp[0][i] = -123456789;
  for (int i = 1; i <= n; i++) {
    int pos1 = i, pos2 = m;
    while (pos1 && pos2) {
      if (s[pos1] == p[pos2]) pos2--;
      pos1--;
    }
    for (int j = 0; j <= n; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      if (!pos2 && j >= i - pos1 - m)
        dp[i][j] = max(dp[i][j], dp[pos1][j - (i - pos1 - m)] + 1);
    }
  }
  for (int i = 0; i <= n; i++) cout << dp[n][i] << " ";
  return 0;
}
