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 long double PI = 3.1415926535897932384626433832795L;
const long double EPS = 1e-9;
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(const long long a, const long long b) {
  return a / gcd(a, b) * b;
}
long long sign(const long long x) { return (x > 0) - (x < 0); }
template <class T>
const T sqr(const T x) {
  return x * x;
}
template <class T>
const T binpow(const T a, const unsigned long long n, const T modulo) {
  return n == 0 ? 1
                : sqr(binpow(a, n / 2, modulo)) % modulo * (n % 2 ? a : 1) %
                      modulo;
}
const int MAXS = 2000;
int min_length_p[MAXS], dp[MAXS + 1][MAXS + 1];
int main() {
  ios::sync_with_stdio(false);
  string s, p;
  cin >> s >> p;
  for (int i = 0; i < s.size(); i++) {
    int s_position = i, p_position;
    for (p_position = 0; s_position < s.size() && p_position < p.size();
         ++s_position) {
      p_position += s[s_position] == p[p_position];
    }
    if (p_position == p.size()) {
      min_length_p[i] = s_position - i;
    } else {
      min_length_p[i] = -1;
    }
  }
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < s.size(); i++) {
    for (int j = 0; j <= i; j++) {
      dp[i + 1][j] = max(dp[i][j], dp[i + 1][j]);
      dp[i + 1][j + 1] = max(dp[i][j], dp[i + 1][j + 1]);
      if (min_length_p[i] != -1) {
        dp[i + min_length_p[i]][j + min_length_p[i] - p.size()] =
            max(dp[i][j] + 1,
                dp[i + min_length_p[i]][j + min_length_p[i] - p.size()]);
      }
    }
  }
  for (int j = 0; j <= s.size(); j++) {
    cout << dp[s.size()][j] << ' ';
  }
  return 0;
}
