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;
char s[2100], pat[555];
int n, m, i, j, k;
int dp[2100][2100];
int main() {
  do {
    s[++n] = getchar();
  } while (s[n] != '\n');
  do {
    pat[++m] = getchar();
  } while (pat[m] != '\n');
  s[n] = '\0';
  pat[m] = '\0';
  n--, m--;
  for (i = 1; i <= n; i++) {
    k = m;
    for (j = i; j && k; j--)
      if (s[j] == pat[k]) k--;
    if (k == 0) {
      int extra = i - j - m;
      for (k = extra; k <= i - m; k++)
        dp[i][k] = max(dp[i][k], dp[j][k - extra] + 1);
    }
    for (k = 0; k < i; k++) dp[i][k] = max(dp[i - 1][k], dp[i][k]);
  }
  for (int i = 0; i < n; i++) cout << dp[n][i] << " ";
  cout << "0" << endl;
}
