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 long long N = 2100;
long long dp[N][N], nxt[N];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s, p;
  cin >> s >> p;
  long long n = s.size(), m = p.size();
  for (long long i = 0; i < n; i++) {
    long long id = 0, j;
    for (j = i; j < n; j++)
      if (s[j] == p[id]) {
        id++;
        if (id == m) break;
      }
    if (id == m)
      nxt[i] = j;
    else
      nxt[i] = -1;
  }
  for (long long i = 0; i <= n; i++) {
    for (long long j = 0; j <= i; j++) {
      if (i != 0) dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (j != 0 && i != 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      if (nxt[i] != -1 && i != n)
        dp[nxt[i] + 1][j + nxt[i] - i + 1 - m] =
            max(dp[nxt[i] + 1][j + nxt[i] - i + 1 - m], dp[i][j] + 1);
    }
  }
  for (long long i = 0; i <= n; i++) cout << dp[n][i] << ' ';
  cout << '\n';
  return 0;
}
