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[3333], p[1111];
int del[3333];
int dp[3333][3333];
int main() {
  cin >> s >> p;
  int n = strlen(s), m = strlen(p);
  for (int i = 0; i < n; i++) {
    int k = i, j = m - 1;
    while (j >= 0 && k >= 0) {
      if (s[k] == p[j]) j--;
      k--;
    }
    if (j == -1)
      del[i + 1] = k + 1;
    else
      del[i + 1] = -1;
  }
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++)
    if (del[i] != -1) {
      int pre = del[i], add = i - del[i] - m;
      for (int j = 0; j <= n; j++) {
        dp[i][j + add] = max(dp[i][j + add], dp[pre][j] + 1);
        dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      }
    }
  for (int i = 0; i <= n; i++) {
    cout << min((n - i) / m, dp[n][i]) << (i == n ? '\n' : ' ');
  }
}
