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][2005];
int nxt[2005];
string s, t;
int n, m;
int solve(int idx, int need) {
  if (need < 0) return -5000;
  if (n - idx < need) return -5000;
  if (idx == n) return 0;
  if (nxt[idx] == -1) return 0;
  int &ret = dp[idx][need];
  if (~ret) return ret;
  ret = solve(idx + 1, need);
  ret = max(ret, solve(idx + 1, need - 1));
  ret = max(ret, 1 + solve(nxt[idx], need - (nxt[idx] - idx - m)));
  return ret;
}
int main() {
  cin >> s >> t;
  n = s.size();
  m = t.size();
  for (int i = 0; i < n; i++) {
    int j = i, k = 0;
    while (j < n && k < m) {
      if (s[j] == t[k]) k++;
      j++;
    }
    nxt[i] = (k == m ? j : -1);
  }
  memset(dp, -1, sizeof dp);
  for (int i = 0; i <= n; i++) cout << solve(0, i) << (i == n ? '\n' : ' ');
  return 0;
}
