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 long long MAXN = 4e3 + 10, inf = 1e18, mod = 1e9 + 7;
struct pair_hash {
  template <class T1, class T2>
  std::size_t operator()(const std::pair<T1, T2> &p) const {
    auto h1 = std::hash<T1>{}(p.first);
    auto h2 = std::hash<T2>{}(p.second);
    return h1 ^ h2;
  }
};
long long pw(long long te, long long tee) {
  if (tee == 0) return 1;
  long long res = pw(te, tee / 2);
  return (tee % 2 ? te * res * res : res * res);
}
string s, p;
long long dp[MAXN][MAXN], a[MAXN], n, m;
void Read_input() { cin >> s >> p; }
void fill_a() {
  for (long long i = 0; i < n; i++) {
    long long k = i, j = 0;
    while (j < m) {
      while (k < n and s[k] != p[j]) k++;
      if (k == n) break;
      k++;
      j++;
    }
    if (j != m)
      a[i] = inf;
    else
      a[i] = k - i;
  }
}
void fill_dp() {
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j <= i; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (a[i] != inf)
        dp[i + a[i]][j + a[i] - m] =
            max(dp[i + a[i]][j + a[i] - m], dp[i][j] + 1);
    }
  }
}
void Solve() {
  n = s.size(), m = p.size();
  fill_a();
  fill_dp();
}
void WriteOut() {
  for (long long i = 0; i <= n; i++) cout << dp[n][i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Read_input(), Solve(), WriteOut();
  return 0;
}
