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;
int main() {
  long long i, j, k;
  string s, p;
  cin >> s >> p;
  long long n = s.length();
  long long dp[n + 1][n + 1];
  vector<long long> a(n);
  for (i = (0); i < (n); i++) {
    long long ind = i;
    for (j = (0); j < (p.length()); j++) {
      while (ind < s.length() && s[ind] != p[j]) ind++;
      if (ind == s.length())
        break;
      else
        ind++;
    }
    if (j == p.length())
      a[i] = (ind - i);
    else
      a[i] = -1;
  }
  for (i = (0); i < (n + 1); i++)
    for (j = (0); j < (n + 1); j++) dp[i][j] = 0;
  for (i = (0); i < (n); i++) {
    for (j = (0); j < (i + 1); 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] != -1) {
        dp[i + a[i]][j + a[i] - p.length()] =
            max(dp[i + a[i]][j + a[i] - p.length()], dp[i][j] + 1);
      }
    }
  }
  for (i = (0); i < (n + 1); i++) cout << dp[n][i] << " ";
  cout << '\n';
  return 0;
}
