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 pre[2005], n, m;
string s, p;
void prec() {
  memset(pre, -1, sizeof pre);
  for (int st = 0; st < n; ++st) {
    int j = 0;
    for (int i = st; i < n; i++) {
      if (s[i] == p[j]) j++;
      if (j == m) {
        pre[st] = i;
        break;
      }
    }
  }
}
int memo[2005][2005];
int solve(int idx, int cnt) {
  if (cnt < 0) return -1e9;
  if (idx == n) return (cnt != 0) * -1e9;
  int &ret = memo[idx][cnt];
  if (~ret) return ret;
  ret = max(solve(idx + 1, cnt), solve(idx + 1, cnt - 1));
  int e = pre[idx];
  if (e != -1) ret = max(ret, solve(e + 1, cnt - (e - idx + 1 - m)) + 1);
  return ret;
}
int main() {
  cin >> s >> p;
  n = s.size(), m = p.size();
  prec();
  memset(memo, -1, sizeof memo);
  for (int i = 0; i < n + 1; ++i) cout << solve(0, i) << " ";
}
