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 = 1005;
int dp[2005][2005];
vector<vector<int> > v;
int nxt[2005];
string s, p;
int calc(int i, int rem) {
  if (rem < 0) return -1e9;
  if (i == s.size()) {
    if (rem == 0)
      return 0;
    else
      return -1e9;
  }
  int &ret = dp[i][rem];
  if (ret != -1) return ret;
  ret = calc(i + 1, rem);
  ret = max(ret, calc(i + 1, rem - 1));
  if (nxt[i] != -1)
    ret = max(ret, 1 + calc(nxt[i] + 1, rem - (nxt[i] - i + 1 - p.size())));
  return ret;
}
int main() {
  cin >> s >> p;
  v.resize(28);
  for (int i = 0; i < s.size(); ++i) {
    v[s[i] - 'a'].push_back(i);
  }
  memset(dp, -1, sizeof dp);
  memset(nxt, -1, sizeof nxt);
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == p[0]) {
      int c = 1;
      int idx = i;
      while (c < p.size() && idx < s.size()) {
        if (upper_bound(v[p[c] - 'a'].begin(), v[p[c] - 'a'].end(), idx) ==
            v[p[c] - 'a'].end())
          break;
        idx = *upper_bound(v[p[c] - 'a'].begin(), v[p[c] - 'a'].end(), idx);
        c++;
      }
      if (c == p.size()) {
        nxt[i] = idx;
      }
    }
  }
  for (int i = 0; i <= s.size(); ++i) {
    cout << calc(0, i) << " ";
  }
  return 0;
}
