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;
string s, p;
int dp[2001][2001];
int j[2001], k[2001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s >> p;
  int n = s.length(), m = p.length();
  for (int i = 0; i < n; i++) {
    if (s[i] == p[m - 1]) {
      bool ok = false;
      for (int a = m - 1, b = i; a >= 0 and b >= 0 and !ok; b--) {
        if (s[b] == p[a])
          a--;
        else
          k[i + 1]++;
        if (a < 0) {
          j[i + 1] = b;
          ok = true;
        }
      }
      if (!ok) k[i + 1] = j[i + 1] = -1;
    } else {
      k[i + 1] = j[i + 1] = -1;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (j[i] != -1 and k[i] == 0) {
      dp[0][i] = dp[0][j[i]] + 1;
    }
    dp[0][i] = max(dp[0][i], dp[0][i - 1]);
  }
  for (int i = 1; i <= n; i++) {
    dp[i][0] = -10000;
    for (int l = 1; l <= n; l++) {
      if (l < i)
        dp[i][l] = -10000;
      else if (l > i and j[l] != -1 and k[l] <= i) {
        dp[i][l] =
            max(dp[i - k[l]][j[l]] + 1, max(dp[i - 1][l - 1], dp[i][l - 1]));
      } else {
        dp[i][l] = max(dp[i - 1][l - 1], dp[i][l - 1]);
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    int mx = 0;
    for (int l = 1; l <= n; l++) {
      mx = max(mx, dp[i][l]);
    }
    cout << mx << " ";
  }
  cout << endl;
  return 0;
}
