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 = 2010;
string a, p;
int ans[N], DP[N][N];
int n, m, k, u, cur;
int maxi(int &u, int v) { u = max(u, v); }
int main() {
  getline(cin, a);
  getline(cin, p);
  n = a.length();
  m = p.length();
  a = " " + a;
  p = " " + p;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) DP[i][j] = (int)(1e9);
  DP[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    k = m;
    u = -1;
    for (int j = i; j >= 1; j--) {
      if (a[j] == p[k]) k--;
      if (k == 0) {
        u = j - 1;
        break;
      }
    }
    for (int j = 0; j <= n / m; j++) {
      DP[i][j] = DP[i - 1][j];
      if (j > 0 && u != -1) DP[i][j] = min(DP[i][j], DP[u][j - 1] + i - u - m);
    }
  }
  for (int i = 1; i <= n / m; i++) {
    for (int j = DP[n][i]; j <= n - i * m; j++) ans[j] = max(ans[j], i);
  }
  for (int i = 0; i <= n; i++) cout << ans[i] << " ";
}
