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 MAXN = 2010;
int dp[MAXN][MAXN], x[MAXN];
int main() {
  string s, p, rs;
  int n;
  cin >> s >> p;
  n = s.size();
  rs = p;
  reverse(rs.begin(), rs.end());
  for (int i = 0; i < n; i++) {
    int pnt = 0;
    if (s[i] != rs[0])
      x[i] = -1;
    else
      for (int j = i; j >= 0; j--) {
        if (s[j] == rs[pnt]) {
          pnt++;
          if (pnt == p.size()) {
            x[i] = j;
            break;
          }
        }
      }
    if (pnt != p.size()) x[i] = -1;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= i; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      if (x[i - 1] != -1 && j - (i - x[i - 1] - p.size()) <= x[i - 1]) {
        dp[i][j] =
            max(dp[i][j], dp[x[i - 1]][j - (i - x[i - 1] - p.size())] + 1);
      }
    }
  for (int j = 0; j <= n; j++) cout << dp[n][j] << " ";
}
