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 = 2505;
int dp[N][N];
int ans[N];
int n, m;
string s, p;
void precal() {
  n = s.size();
  m = p.size();
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 1;
  if (s[0] == p[0]) dp[0][1] = 1;
  for (int i = 1; i < n; ++i) {
    dp[i][0] = i + 1;
    for (int j = 1; j <= i + 1; ++j) {
      if (s[i] == p[(j + m - 1) % m] && dp[i - 1][j - 1] != -1)
        dp[i][j] = dp[i - 1][j - 1] + 1;
      if (dp[i - 1][j] != -1)
        dp[i][j] = max(dp[i][j], dp[i - 1][j] + (j % m == 0));
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (dp[n - 1][i] == -1) continue;
    ans[dp[n - 1][i]] = max(ans[dp[n - 1][i]], i);
  }
  for (int i = n; i >= 0; --i) ans[i] = max(ans[i], ans[i + 1]);
  for (int i = n; i >= 0; --i) {
    cout << min(i, ans[i]) / m << ' ';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> p;
  precal();
}
