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 int maxn = 2010;
int dp[maxn][maxn];
char s[maxn], p[maxn];
int main() {
  scanf("%s%s", s + 1, p + 1);
  int n1 = strlen(s + 1), n2 = strlen(p + 1);
  memset(dp, 0, sizeof(dp));
  for (int i = n2; i <= n1; i++) {
    int k = n2, t = i;
    while (t >= 1) {
      if (s[t] == p[k]) k--;
      if (k == 0) break;
      t--;
    }
    if (k == 0) {
      for (int j = 0; j <= i; j++)
        if (j >= i - t + 1 - n2 && t - 1 >= j + n2 - i + t - 1)
          dp[i][j] = dp[t - 1][j + n2 - i + t - 1] + 1;
    }
    for (int j = 0; j <= i; j++) dp[i][j] = max(dp[i][j], dp[i - 1][j]);
  }
  for (int i = 0; i < n1; i++) cout << dp[n1][i] << " ";
  cout << dp[n1][n1] << endl;
  return 0;
}
