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 int N = 2000 + 5;
int n, m, dp[N][N];
string s, p;
void readInput() {
  cin >> s >> p;
  n = s.size();
  m = p.size();
}
void solve() {
  memset(dp, -63, sizeof dp);
  for (int i = dp[0][0] = 0; i < n; i++) {
    int nxt = i, cur = m - 1;
    for (int j = i; ~j && ~cur; j--)
      if (s[j] == p[cur]) cur--, nxt = j - 1;
    if (~cur) nxt = -2;
    for (int j = 0; j <= i + 1; j++)
      dp[i + 1][j] = max(
          max(j ? dp[i][j - 1] : 0, dp[i][j]),
          nxt ^ -2 && j >= i - nxt - m ? dp[nxt + 1][j - i + nxt + m] + 1 : 0);
  }
}
void writeOutput() {
  for (int i = 0; i <= n; i++) cout << dp[n][i] << ' ';
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  readInput(), solve(), writeOutput();
  return 0;
}
