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 a[2010], dp[2010][2010];
int main() {
  string s, p;
  int i, j, k, ns, np;
  cin >> s >> p;
  ns = s.length();
  np = p.length();
  for (i = 0; i < ns; i++) {
    j = 0;
    k = i;
    while (j < np && k < ns) {
      if (p[j] == s[k]) j++;
      k++;
    }
    a[i] = j == np ? k - i : -1;
  }
  for (i = 0; i < ns; i++)
    for (j = 0; j <= i; j++) {
      dp[i + 1][j] = max(dp[i][j], dp[i + 1][j]);
      dp[i + 1][j + 1] = max(dp[i][j], dp[i + 1][j + 1]);
      if (a[i] > 0)
        dp[i + a[i]][j + a[i] - np] =
            max(dp[i][j] + 1, dp[i + a[i]][j + a[i] - np]);
    }
  i = ns;
  cout << dp[i][0];
  for (j = 1; j <= ns; j++) cout << " " << dp[i][j];
}
