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;
char s[2005], p[505];
int n, m;
vector<vector<int> > dp[2005];
int go(int idx, int made, int p_idx) {
  if (made == 0) return 0;
  int &ret = dp[idx][made][p_idx];
  if (ret != -1) return ret;
  ret = 10000;
  if (idx == n) return ret = 10000;
  if (p_idx == 0) ret = min(ret, go(idx + 1, made, p_idx));
  if (s[idx] == p[p_idx]) {
    if (p_idx == m - 1)
      ret = min(ret, go(idx + 1, made - 1, 0));
    else
      ret = min(ret, go(idx + 1, made, p_idx + 1));
  }
  ret = min(ret, 1 + go(idx + 1, made, p_idx));
  return ret;
}
int main() {
  scanf("%s%s", s, p);
  n = strlen(s), m = strlen(p);
  for (int i = 0; i <= n + 1; ++i) {
    dp[i] = vector<vector<int> >(n / m + 5, vector<int>(m + 5, -1));
  }
  for (int x = 0; x <= n; ++x) {
    int ans = 0;
    for (int made = 0; made <= n / m; ++made) {
      int r = go(0, made, 0);
      if (r > x) continue;
      int rest = n - r - made * m;
      int erase = max(0, x - r - rest);
      ans = max(ans, made - (erase + m - 1) / m);
    }
    printf("%d ", ans);
  }
}
