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 = 2005;
const int P = 505;
char s[N], p[N];
int n, m;
int pos[N];
int dp[N][N];
int f(int i, int j) {
  if (j < 0) return -1e9;
  if (i == n) return j == 0 ? 0 : -1e9;
  int &ans = dp[i][j];
  if (~ans) return ans;
  ans = -1e9;
  ans = max(ans, f(i + 1, j));
  ans = max(ans, f(i + 1, j - 1));
  if (pos[i] < n) {
    ans = max(ans, 1 + f(pos[i] + 1, j - (pos[i] - i + 1 - m)));
  }
  return ans;
}
int main() {
  memset(dp, -1, sizeof dp);
  scanf("%s %s", s, p);
  n = (int)strlen(s);
  m = (int)strlen(p);
  for (int i = 0; i < n; i++) {
    int k = 0;
    pos[i] = n;
    for (int j = i; j < n; j++) {
      if (s[j] == p[k]) k++;
      if (k == m) {
        pos[i] = j;
        break;
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    printf("%d%c", f(0, i), " \n"[i == n]);
  }
}
