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 t, p;
int dp[2009][2009], arr[2009], n, m;
int memoize(int idx, int rem) {
  if (idx == n && rem == 0) return 0;
  if (idx == n || rem < 0) return -1 * 2000000000;
  if (dp[idx][rem] != -1) return dp[idx][rem];
  int ans = -1 * 2000000000;
  ans = max(ans, memoize(idx + 1, rem));
  ans = max(ans, memoize(idx + 1, rem - 1));
  if (arr[idx] != -1)
    ans = max(ans, 1 + memoize(arr[idx] + 1, rem - (arr[idx] - idx + 1 - m)));
  return dp[idx][rem] = ans;
}
void pre() {
  n = t.length();
  m = p.length();
  for (int i = int(0); i <= int(n); i++) arr[i] = -1;
  for (int i = int(0); i <= int(n - 1); i++) {
    int cnt = 0;
    for (int j = int(i); j <= int(n - 1); j++) {
      if (t[j] == p[cnt]) cnt++;
      if (cnt == m) {
        arr[i] = j;
        break;
      }
    }
  }
}
int main() {
  cin >> t >> p;
  pre();
  memset(dp, -1, sizeof(dp));
  for (int i = int(0); i <= int(n); i++) {
    printf("%d ", memoize(0, i));
  }
  printf("\n");
  return 0;
}
