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 nmax = 2e3 + 42, inf = 1e9;
string s, p;
int n;
int where[nmax];
int dp[nmax][nmax];
int outp[nmax];
int rec(int pos, int rem) {
  if (rem == 0) return 0;
  if (pos == n) return inf;
  if (dp[pos][rem] != -1) return dp[pos][rem];
  int ans = rec(pos + 1, rem);
  if (where[pos] != -1) {
    int v = rec(where[pos], rem - 1) + where[pos] - pos - p.size();
    ans = min(ans, v);
  }
  dp[pos][rem] = ans;
  return ans;
}
int main() {
  cin >> s >> p;
  n = s.size();
  for (int i = 0; i < n; i++) {
    int now = 0;
    int j = i;
    while (now < p.size() && j < n) {
      if (p[now] == s[j]) now++;
      j++;
    }
    if (now == p.size())
      where[i] = j;
    else
      where[i] = -1;
  }
  memset(dp, -1, sizeof(dp));
  int pos = 0, t = 0;
  while (pos < n) {
    if (where[pos] != -1)
      pos = where[pos], t++;
    else
      break;
  }
  for (int c = 1; c <= t; c++) {
    for (int j = rec(0, c); j <= n - c * p.size(); j++)
      outp[j] = max(outp[j], c);
  }
  for (int i = 0; i <= n; i++) cout << outp[i] << " ";
  cout << endl;
  return 0;
}
