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 = 2e3 + 5;
string s, p;
int nxt[MAX];
vector<int> cost;
int memo[MAX][MAX];
int roll(int idx, int val) {
  if (idx >= (int)cost.size()) {
    return 0;
  }
  int &ans = memo[idx][val];
  if (~ans) {
    return ans;
  }
  ans = roll(idx + 1, val);
  if (val >= (int)cost[idx]) {
    ans = max(ans, roll(nxt[idx], val - cost[idx]) + 1);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> s >> p;
  vector<pair<int, int> > ranges;
  for (int i = int(0); i < int(s.size()); i++) {
    int idx = 0;
    int match = -1;
    if (s[i] == p[0]) {
      for (int j = int(i); j < int(s.size()); j++) {
        if (p[idx] == s[j]) {
          idx++;
        }
        if (idx == (int)p.size()) {
          match = j;
          break;
        }
      }
    }
    if (match != -1) {
      ranges.emplace_back(i, match);
    }
  }
  for (int i = int(0); i < int(ranges.size()); i++) {
    auto it = lower_bound(ranges.begin(), ranges.end(),
                          make_pair(ranges[i].second + 1, 0));
    nxt[i] = it - ranges.begin();
    cost.push_back(
        max(0, ranges[i].second - ranges[i].first - (int)p.size() + 1));
  }
  memset(memo, -1, sizeof memo);
  for (int i = int(0); i < int(s.size() + 1); i++) {
    cout << min(roll(0, i), ((int)s.size() - i) / (int)p.size()) << " ";
  }
  cout << '\n';
  return 0;
}
