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>
static const int INF = 1000000000;
class C {
 public:
  void solve(std::istream& in, std::ostream& out) {
    std::string s, p;
    in >> s >> p;
    std::vector<int> go(s.size() + 1, 0);
    for (int i = 0; i <= s.size(); ++i) {
      int pos = i;
      for (int j = 0; j < p.size(); ++j) {
        while (pos != s.size() && s[pos] != p[j]) {
          ++pos;
        }
        if (pos == s.size()) {
          pos = INF;
          break;
        } else
          ++pos;
      }
      go[i] = pos;
    }
    std::vector<std::vector<int>> dp(s.size() + 1,
                                     std::vector<int>(s.size() + 2, -INF));
    dp[0][0] = 0;
    for (int i = 0; i < s.size(); ++i) {
      for (int j = 0; j <= s.size(); ++j) {
        if (dp[i][j] < 0) continue;
        {
          int newI = i + 1;
          int newJ = j + 1;
          dp[newI][newJ] = std::max(dp[newI][newJ], dp[i][j]);
        }
        {
          int newI = i + 1;
          int newJ = j;
          dp[newI][newJ] = std::max(dp[newI][newJ], dp[i][j]);
        }
        {
          int newI = go[i];
          if (go[i] == INF) continue;
          int newJ = (int)(j + go[i] - i - p.size());
          dp[newI][newJ] = std::max(dp[newI][newJ], dp[i][j] + 1);
        }
      }
    }
    for (int i = 0; i <= s.size(); ++i) {
      out << dp[s.size()][i] << " ";
    }
  }
};
int main() {
  C solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
