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;
long long a[2005], dp[2005][2005];
string second, p;
long long first(long long ind) {
  long long ind2 = 0;
  while (ind < (long long)second.size() && ind2 < (long long)p.size()) {
    if (second[ind] == p[ind2]) ind2++;
    ind++;
    if (ind2 == (long long)p.size()) return ind;
  }
  return -1;
}
void calc() {
  for (long long i = 0; i < (long long)second.size(); i++) a[i] = first(i);
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> second >> p;
  calc();
  for (long long i = 0; i < (long long)second.size(); i++)
    for (long long j = 0; j <= i; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (a[i] != -1)
        dp[a[i]][j + a[i] - (long long)p.size() - i] =
            max(dp[a[i]][j + a[i] - (long long)p.size() - i], dp[i][j] + 1);
    }
  for (long long i = 0; i <= (long long)second.size(); i++)
    cout << dp[(long long)second.size()][i] << " ";
  return 0;
}
