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 pre[2020];
int n, m;
string s, t;
int dp[2020][2020];
int f(int i, int c) {
  if (i == n) {
    if (c == 0) return 0;
    return -1e4;
  }
  if (dp[i][c] != -1) return dp[i][c];
  int r1 = -1e4;
  if (s[i] == t[0] && c - m >= 0 && pre[i] != -1) r1 = 1 + f(pre[i], c - m);
  if (c) r1 = max(r1, f(i + 1, c - 1));
  r1 = max(r1, f(i + 1, c));
  return dp[i][c] = r1;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> s >> t;
  n = s.length();
  m = t.length();
  for (int i = 0; i < n; i++) {
    int k = 0;
    for (int j = i; j < n; j++) {
      if (s[j] == t[k]) k++;
      if (k == m) {
        pre[i] = j + 1;
        break;
      }
    }
    if (!pre[i]) pre[i] = -1;
  }
  memset(dp, -1, sizeof dp);
  for (int i = 0; i <= n; i++) cout << f(0, n - i) << " ";
  cout << endl;
  return 0;
}
