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;
int dp[2005][2004];
string str, str2;
int i, j;
int f(int x, int t) {
  if (t == -1) return 0;
  int &r = dp[x][t];
  if (r != -1) return r;
  if (x == str.size()) return 1000000007;
  r = f(x + 1, t);
  int a = str2.size() - t % str2.size() - 1;
  if (a) r++;
  if (str[x] == str2[a]) r = min(f(x + 1, t - 1), r);
  return r;
}
int main() {
  memset(dp, -1, sizeof dp);
  cin >> str >> str2;
  for (i = 1; i <= str.size() / str2.size(); i++) f(0, i * str2.size() - 1);
  for (i = 0; i <= str.size(); i++) {
    for (j = str.size() / str2.size(); j >= 1; j--) {
      int t = dp[0][j * str2.size() - 1];
      int p = j * str2.size();
      if (i >= t and i <= str.size() - p) break;
    }
    cout << j << ' ';
  }
}
