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 MAX = 2 * 1000 + 5;
const int MAXN = 505;
pair<int, int> dp[MAX];
int d[MAX][MAX];
int main() {
  string s, p;
  cin >> s >> p;
  for (int i = 0; i <= MAX; i++) dp[i].first = dp[i].second = -1;
  for (int i = 0; i < s.size(); i++) {
    int ind1 = i;
    int ind2 = 0;
    int cnt = 0;
    while (ind2 != p.size() and ind1 != s.size()) {
      if (s[ind1] == p[ind2]) {
        ind2++;
        ind1++;
      } else {
        cnt++;
        ind1++;
      }
    }
    if (ind2 == p.size()) {
      dp[i].second = cnt;
      dp[i].first = ind1 - i;
    }
  }
  for (int i = 0; i < s.size(); i++) {
    for (int j = 0; j <= i; j++) {
      d[i + 1][j + 1] = max(d[i][j], d[i + 1][j + 1]);
      d[i + 1][j] = max(d[i][j], d[i + 1][j]);
      if (dp[i].first != -1) {
        d[i + dp[i].first][j + dp[i].second] =
            max(d[i + dp[i].first][j + dp[i].second], d[i][j] + 1);
      }
    }
  }
  for (int j = 0; j <= s.size(); j++) {
    cout << d[s.size()][j] << " ";
  }
  cout << endl;
}
