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;
long long mod = 1000000007;
int n;
using namespace std;
string s, p;
int dp[2005][2005];
pair<int, int> use[2005];
int get(int idx, int rem) {
  if (rem < 0) return -2000000000;
  if (dp[idx][rem] != -1) return dp[idx][rem];
  int orex = idx;
  int orem = rem;
  dp[orex][orem] = max(get(idx + 1, rem), get(idx + 1, rem - 1));
  if (use[orex].first <= orem) {
    idx = use[orex].second;
    rem -= use[orex].first;
    dp[orex][orem] = max(dp[orex][orem], 1 + get(idx, rem));
  }
  return dp[orex][orem];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> p;
  for (int i = 0; i < 2005; i++)
    for (int j = 0; j < 2005; j++) dp[i][j] = -1;
  for (int i = 0; i < s.length() + 1; i++) dp[s.length()][i] = -2000000000;
  dp[s.length()][0] = 0;
  for (int j = 0; j < s.length(); j++) {
    int idx = j;
    int rem = 0;
    for (int i = 0; i < p.length();) {
      if (idx >= s.length()) {
        idx = 2000000000;
        use[j] = make_pair(2000000000, 2000000000);
        break;
      }
      if (s[idx] == p[i]) {
        i++;
        idx++;
      } else {
        idx++;
        rem++;
      }
    }
    if (idx <= s.length()) {
      use[j] = make_pair(rem, idx);
    }
  }
  cout << get(0, 0);
  for (int i = 1; i <= s.length(); i++) cout << ' ' << get(0, i);
  cout << endl;
  return 0;
}
