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;
string s, p;
const long long MAXN = 2e3 + 10;
long long dp[MAXN][MAXN], d;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> s >> p;
  for (long long i = 0; i < (long long)s.size(); i++) {
    long long o = p.size() - 1;
    if (s[i] == p[o]) {
      for (d = i; d >= 0; d--) {
        if (s[d] == p[o]) o--;
        if (o < 0) break;
      }
    }
    long long u = i - d + 1 - p.size();
    for (long long j = 0; j <= i + 1; j++) {
      dp[i][j] = 0;
      if (i) dp[i][j] = dp[i - 1][j];
      if (i && j) dp[i][j] = max(dp[i - 1][j - 1], dp[i][j]);
      if (o < 0) {
        if (j >= u) {
          if (d) {
            if (j - u <= d) dp[i][j] = max(dp[i][j], dp[d - 1][j - u] + 1);
          } else if (j == u)
            dp[i][j] = max(dp[i][j], (long long)1);
        }
      }
    }
  }
  for (long long i = 0; i <= (long long)s.size(); i++)
    cout << dp[(long long)s.size() - 1][i] << " ";
}
