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;
int recur_depth = 0;
const long long sz = 2009;
long long dp[sz][sz], nxt[sz], rmv[sz];
string s, p;
long long solve(long long pos, long long x) {
  if (x < 0) return -(1LL << 61);
  if (pos == s.size()) {
    if (x == 0)
      return 0;
    else
      return -(1LL << 61);
  }
  long long& ret = dp[pos][x];
  if (ret != -1) return ret;
  ret = max(solve(pos + 1, x), solve(pos + 1, x - 1));
  if (nxt[pos] != -1) ret = max(ret, 1 + solve(nxt[pos], x - rmv[pos]));
  return ret;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> s >> p;
  for (long long i = 0; i < s.size(); i++) {
    long long cnt = 0, cur = i, ptr = 0;
    while (cur < s.size() && ptr < p.size()) {
      if (s[cur] == p[ptr])
        ptr++;
      else
        cnt++;
      cur++;
    }
    if (ptr != p.size())
      nxt[i] = -1;
    else
      rmv[i] = cnt, nxt[i] = cur;
  }
  memset(dp, -1, sizeof(dp));
  for (long long x = 0; x <= s.size(); x++) cout << solve(0, x) << " ";
  cout << '\n';
  return 0;
}
