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 oo = 2014;
string s, p;
int n, m, lp[oo][oo], dp[oo][oo];
int lastPos(int sidx, int pidx) {
  if (pidx == m) {
    return sidx;
  }
  if (sidx == n) {
    return oo;
  }
  if (lp[sidx][pidx] != -1) {
    return lp[sidx][pidx];
  }
  int ans = oo;
  if (s[sidx] == p[pidx]) {
    ans = min(ans, lastPos(sidx + 1, pidx + 1));
  } else {
    ans = min(ans, lastPos(sidx + 1, pidx));
  }
  return lp[sidx][pidx] = ans;
}
int f(int idx, int toBeDeleted) {
  if (idx == n) {
    if (toBeDeleted) {
      return -oo;
    } else {
      return 0;
    }
  }
  if (dp[idx][toBeDeleted] != -1) {
    return dp[idx][toBeDeleted];
  }
  int ans = max(f(idx + 1, toBeDeleted), f(idx + 1, toBeDeleted - 1));
  int pos = lastPos(idx, 0);
  int deleted = pos - idx - m;
  if (pos != oo and toBeDeleted >= deleted) {
    ans = max(ans, f(pos, toBeDeleted - deleted) + 1);
  }
  return dp[idx][toBeDeleted] = ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> s >> p;
  n = s.size(), m = p.size();
  for (int i = 0; i < oo; i++) {
    for (int j = 0; j < oo; j++) {
      dp[i][j] = lp[i][j] = -1;
    }
  }
  for (int i = 0; i <= n; i++) {
    cout << f(0, i) << " ";
  }
  cout << "\n";
}
