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 = 2007;
const int inf = 1e9 + 7;
string s, t;
int n, m;
int dp[N][N];
int find(int i) {
  int j = m;
  while (j > 0 && i > 0) {
    if (s[i - 1] == t[j - 1]) j--;
    i--;
  }
  if (j > 0) return -1;
  return i;
}
int main() {
  getline(cin, s);
  getline(cin, t);
  n = s.size();
  m = t.size();
  for (int i = 1; i <= n; i++) {
    int t = find(i);
    if (t >= 0 && i - t == m)
      dp[i][0] = max(dp[i - 1][0], dp[t][0] + 1);
    else
      dp[i][0] = dp[i - 1][0];
    for (int j = 1; j < i; j++) {
      dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][j]);
      int tmp = i - t - m;
      if (t >= 0 && t >= j - tmp && j >= tmp)
        dp[i][j] = max(dp[i][j], dp[t][j - tmp] + 1);
    }
    dp[i][i] = 0;
  }
  for (int i = 0; i <= n; i++) cout << dp[n][i] << " ";
}
