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 MAXN = 2010, MAXM = 510, INF = 1e9;
int dp[MAXN][MAXN];
string s, p;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> s >> p;
  int n = s.size(), m = p.size();
  for (int i = 0; i < n; ++i) {
    int t = m, k = i + 1;
    while (t && k--)
      if (s[k] == p[t - 1]) --t;
    --k;
    for (int j = 0; j <= i; ++j) {
      if (i) dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (i && j) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      if (!t && j >= i - k - m && k - (j - (i - k - m)) + 1 >= 0) {
        if (k >= 0)
          dp[i][j] = max(dp[i][j], dp[k][j - (i - k - m)] + 1);
        else
          dp[i][j] = max(dp[i][j], 1);
      }
    }
    for (int j = i + 2; j <= n; ++j) dp[i][j] = -INF;
  }
  for (int i = 0; i <= n; ++i) cout << dp[n - 1][i] << ' ';
  return 0;
}
