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 A[2222];
string str, p;
int dp[2222][2222];
bool done[2222][2222];
int rec(int index, int removed) {
  if (index == str.size())
    if (removed != 0)
      return INT_MIN;
    else
      return 0;
  if (done[index][removed]) return dp[index][removed];
  int ans = INT_MIN;
  if (removed) ans = max(ans, rec(index + 1, removed - 1));
  ans = max(ans, rec(index + 1, removed));
  int remaning = removed - (A[index] - (index + p.size()));
  if (remaning >= 0 && A[index] != -1)
    ans = max(ans, 1 + rec(A[index], remaning));
  done[index][removed] = 1;
  return dp[index][removed] = ans;
}
int main() {
  cin >> str >> p;
  for (int i = 0; i < str.size(); i++) {
    int j;
    int k = i;
    for (j = 0; j < p.size() && k < str.size();) {
      if (str[k] == p[j]) {
        j++;
        k++;
      } else
        k++;
    }
    if (j == p.size())
      A[i] = k;
    else
      A[i] = -1;
  }
  for (int i = 0; i <= str.size(); i++) cout << rec(0, i) << " ";
  return 0;
}
