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;
inline void smax(int &a, int b) {
  if (a < b) a = b;
}
const int N = 2010, A = 26;
string s, p;
int n, m;
int MYNEXT[N][A], comp[N];
int dp[N][N];
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cin >> s >> p;
  n = s.size();
  m = p.size();
  for (int i = 0; i < n; i++) s[i] -= 'a';
  for (int i = 0; i < m; i++) p[i] -= 'a';
  for (int i = 0; i < A; i++) MYNEXT[n][i] = n;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < A; j++) MYNEXT[i][j] = MYNEXT[i + 1][j];
    MYNEXT[i][s[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    int pos = i;
    for (int j = 0; j < m; j++) {
      pos = MYNEXT[pos][p[j]];
      pos++;
      if (pos >= n && j != m - 1) {
        pos = n + 1;
        break;
      }
    }
    comp[i] = pos;
  }
  memset(dp, -63, sizeof dp);
  memset(dp[0], 0, sizeof dp[0]);
  for (int i = 0; i <= n; i++)
    for (int j = n - 1; j >= 0; j--) {
      smax(dp[i][j], dp[i][j + 1]);
      if (i) smax(dp[i][j], dp[i - 1][j + 1]);
      if (comp[j] != n + 1 && comp[j] - j - m <= i)
        smax(dp[i][j], dp[i - (comp[j] - j - m)][comp[j]] + 1);
    }
  for (int i = 0; i <= n; i++) cout << dp[i][0] << " ";
}
