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 = 2050;
string a, b;
int n, m, nxt[N], dp[N][N];
int solve(int i, int j) {
  if (i == n) return (!j ? 0 : -1e9);
  int &x = dp[i][j];
  if (x != -1) return x;
  x = -1e9;
  x = max(x, max(solve(i + 1, j), solve(i + 1, j - 1)));
  if (nxt[i] != -1) x = max(x, 1 + solve(i + nxt[i], j - (nxt[i] - m)));
  return x;
}
int main() {
  cin >> a >> b;
  n = a.size(), m = b.size();
  memset(dp, -1, sizeof dp);
  for (int i = 0; i < n; i++) {
    int j = 0, next = 0;
    for (int k = i; k < n && j < m; k++, next++)
      if (a[k] == b[j]) j++;
    nxt[i] = (j == m ? next : -1);
  }
  for (int i = 0; i <= n; i++) printf("%d ", solve(0, i));
}
