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 dp[2002][2002];
int ri[2002];
string s, p;
vector<int> st;
void nxt(int idx) {
  int i = idx;
  int j = 0;
  int ret = 0;
  while (i < s.size() && j < p.size()) {
    if (s[i] == p[j]) {
      j++;
      i++;
    } else {
      ret++;
      i++;
    }
  }
  if (p.size() == j)
    ri[idx] = i;
  else
    ri[idx] = 1e9;
}
int solve(int i, int k) {
  int &ret = dp[i][k];
  if (ret != -1) return ret;
  ret = 0;
  if (i + k > s.size()) {
    return ret = -1e9;
  }
  if (i + k == s.size()) {
    return ret = 0;
  }
  if (i == s.size()) {
    if (k == 0)
      return ret = 0;
    else
      return ret = -1e9;
  }
  ret = solve(i + 1, k);
  if (k > 0) ret = max(ret, solve(i + 1, k - 1));
  if (s[i] == p[0]) {
    if (ri[i] <= s.size()) {
      int rem = ri[i] - i - p.size();
      if (rem <= k) {
        ret = max(ret, 1 + solve(ri[i], k - rem));
      }
    }
  }
  return ret;
}
int main() {
  cin >> s >> p;
  for (int i = 0; i <= s.size(); i++)
    for (int j = 0; j <= s.size(); j++) dp[i][j] = -1;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == p[0]) {
      st.push_back(i);
      nxt(i);
    }
  }
  for (int i = 0; i <= s.size(); i++) {
    cout << solve(0, i) << ' ';
  }
  cout << endl;
  return 0;
}
