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>
const int N = 2007;
char s[N], p[N];
int nast[N];
int n, m;
void licz_ile(int pos) {
  nast[pos] = -1;
  int t = pos;
  for (int i = 1; i <= m; i++) {
    while (t <= n && s[t] != p[i]) t++;
    if (t > n) return;
    t++;
  }
  nast[pos] = t;
}
int dp[N][N];
int main() {
  std::cin >> (s + 1) >> (p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  for (int i = 1; i <= n; i++) {
    licz_ile(i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      dp[i + 1][j] = std::max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = std::max(dp[i + 1][j + 1], dp[i][j]);
      if (nast[i] != -1) {
        int dz = nast[i] - i - m;
        dp[nast[i]][j + dz] = std::max(dp[nast[i]][j + dz], 1 + dp[i][j]);
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    std::cout << dp[n + 1][i] << " ";
  }
  std::cout << std::endl;
}
