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;
constexpr int N = 2000 + 7;
vector<pair<int, int> > v[N];
int dp[N][N];
int32_t main() {
  ios::sync_with_stdio(0);
  string s, t;
  cin >> s >> t;
  s = '#' + s;
  for (int i = 1; i < ((int)s.size()); i++) {
    int pos = 1, j = i + 1;
    if (s[i] == t[0]) {
      for (; (pos < ((int)t.size())) && (j < ((int)s.size())); j++)
        if (s[j] == t[pos]) pos++;
      if (pos == ((int)t.size()))
        v[j - 1].push_back({i, j - i - ((int)t.size())});
    }
  }
  for (int i = 1; i < ((int)s.size()); i++) {
    for (int j = 0; j <= i; j++) {
      dp[i][j] = max(dp[i - 1][j], j ? dp[i - 1][j - 1] : 0);
      for (auto k : v[i])
        if (j >= k.second && (k.first - 1 >= j - k.second))
          dp[i][j] = max(dp[i][j], dp[k.first - 1][j - k.second] + 1);
    }
  }
  for (int i = 0; i < ((int)s.size()); i++)
    cout << dp[((int)s.size()) - 1][i] << ' ';
}
