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 N = 2e3 + 1;
int dp[N][N];
int main() {
  string s1, s2;
  cin >> s1 >> s2;
  int i2 = 0;
  for (int i = 1; i <= s1.size(); i++) {
    int l2 = s2.size();
    for (i2 = i; i2 && l2; i2--)
      if (s1[i2 - 1] == s2[l2 - 1]) l2--;
    if (!l2) {
      int deletee = i - i2 - s2.size();
      for (int i3 = deletee; i3 <= i - s2.size(); i3++)
        dp[i][i3] = max(dp[i][i3], dp[i2][i3 - deletee] + 1);
    }
    for (i2 = 0; i2 < i; i2++) dp[i][i2] = max(dp[i][i2], dp[i - 1][i2]);
  }
  for (int i = 0; i < s1.size(); i++) cout << dp[s1.size()][i] << " ";
  cout << 0;
  return 0;
}
