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 n, m, dp[2002][2002];
char s[2002], p[2002];
inline int Solve(int i) {
  int j = m;
  for (; i >= 1; --i) {
    if (s[i] == p[j]) --j;
    if (j == 0) return i;
  }
  return -1;
}
int main() {
  cin >> (s + 1) >> (p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= n; ++j) dp[i][j] = -1000000;
  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    int pos = Solve(i);
    int cnt = i - pos + 1 - m;
    for (int j = 0; j <= i; ++j) {
      dp[i][j] = dp[i - 1][j];
      if (j > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      if (pos > 0 && j >= cnt)
        dp[i][j] = max(dp[i][j], dp[pos - 1][j - cnt] + 1);
    }
  }
  for (int i = 0; i <= n; ++i) cout << dp[n][i] << " ";
  cout << "\n";
  return 0;
}
