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;
string str, pat;
int n, p;
long long dp[2234][2234];
long long pos[1234567];
long long recur(int ind, int num) {
  if (ind == 0) {
    if (num == 0) return 0;
    return -123456789;
  }
  if (dp[ind][num] != -1) return dp[ind][num];
  long long ans = -123456789;
  if (num > 0) ans = max(ans, recur(ind - 1, num - 1));
  ans = max(ans, recur(ind - 1, num));
  if (pos[ind] != -1) {
    int x = ((ind - pos[ind] + 1) - p);
    if (x <= num) ans = max(ans, 1 + recur(pos[ind] - 1, num - x));
  }
  return dp[ind][num] = ans;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin >> str;
  cin >> pat;
  n = str.length();
  p = pat.length();
  memset(dp, -1, sizeof(dp));
  memset(pos, -1, sizeof(pos));
  int i;
  for (i = 1; i < n + 1; i++) {
    int ind = i;
    int cnt = 0;
    while (ind > 0 && cnt < p) {
      if (str[ind - 1] == pat[p - 1 - cnt]) cnt++;
      ind--;
    }
    if (cnt == p) pos[i] = ind + 1;
  }
  for (i = 0; i < n + 1; i++) {
    cout << recur(n, i) << " ";
  }
  return 0;
}
