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;
const int MAXL = 510;
const int oo = 1E9;
string s, p;
int dp[MAXN][MAXL], ndp[MAXN][MAXL];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> s >> p;
  int maxs = ((int)(s).size()) / ((int)(p).size());
  for (int i = 0; i <= maxs; ++i)
    for (int j = 0; j < ((int)(p).size()); ++j) dp[i][j] = oo;
  dp[0][0] = 0;
  for (int l = 0; l < ((int)(s).size()); ++l) {
    for (int i = 0; i <= maxs; ++i)
      for (int j = 0; j < ((int)(p).size()); ++j) ndp[i][j] = oo;
    for (int i = 0; i <= maxs; ++i)
      for (int j = 0; j < ((int)(p).size()); ++j)
        if (dp[i][j] < oo) {
          if (s[l] == p[j]) {
            if (j + 1 == ((int)(p).size())) {
              (ndp[i + 1][0]) = min((ndp[i + 1][0]), (dp[i][j]));
            } else {
              (ndp[i][j + 1]) = min((ndp[i][j + 1]), (dp[i][j]));
            }
          }
          (ndp[i][j]) = min((ndp[i][j]), (dp[i][j] + (j != 0 ? 1 : 0)));
        }
    for (int i = 0; i <= maxs; ++i)
      for (int j = 0; j < ((int)(p).size()); ++j) dp[i][j] = ndp[i][j];
  }
  vector<int> ans(((int)(s).size()) + 1, 0);
  for (int i = 1; i <= maxs; ++i)
    for (int j = 0; j < ((int)(p).size()); ++j)
      if (dp[i][j] < oo) {
        int take = dp[i][j];
        int free = ((int)(s).size()) - take - i * ((int)(p).size());
        for (int a = take; a <= ((int)(s).size()); ++a) {
          int left = max(0, a - take - free);
          int give = i - (left + ((int)(p).size()) - 1) / ((int)(p).size());
          (ans[a]) = max((ans[a]), (give));
        }
      }
  for (int i = 0; i <= ((int)(s).size()); ++i)
    printf("%d%c", ans[i], i == ((int)(s).size()) ? '\n' : ' ');
  return 0;
}
