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 N = 2001;
int dp[N][N], n, k;
string s, p;
int getL(int pos) {
  int now = k - 1;
  for (int i = pos; i >= 0; i--)
    if (s[i] == p[now]) {
      if (now == 0) return i;
      now--;
    }
  return -1;
}
int main() {
  cin >> s >> p;
  n = s.size();
  k = p.size();
  for (int i = 0; i < n; i++) {
    int l = getL(i);
    int need = i - l + 1 - k;
    for (int j = 0; j <= i + 1; j++) {
      if (i > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (i > 0 && j > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      if (l != -1 && j >= need && j - need <= l)
        dp[i][j] = max(dp[i][j], (l ? dp[l - 1][j - need] : 0) + 1);
    }
  }
  for (int i = 0; i <= n; i++) cout << dp[n - 1][i] << ' ';
  cout << endl;
  return 0;
}
