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;
int st[2010];
int dp[2010][2010];
int main() {
  string s, p;
  while (cin >> s >> p) {
    for (int i = 0; i < 2010; i++) {
      st[i] = -1;
      for (int j = 0; j < 2010; j++) {
        dp[i][j] = -123456789;
      }
      dp[i][i] = 0;
    }
    for (int i = 0; i < (int)s.length(); i++) {
      for (int j = i, k = p.length() - 1; j >= 0; j--) {
        if (s[j] == p[k]) {
          k--;
        }
        if (k < 0) {
          st[i] = j;
          break;
        }
      }
    }
    for (int i = 1; i <= (int)s.length(); i++) {
      for (int j = 0; j <= i; j++) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j]);
        if (st[i - 1] >= 0) {
          int tmp = i - st[i - 1] - p.length();
          if (tmp <= j) {
            dp[i][j] = max(dp[i][j], dp[st[i - 1]][j - tmp] + 1);
          }
        }
      }
    }
    for (int i = 0; i <= (int)s.length(); i++) {
      cout << dp[s.length()][i] << " ";
    }
    cout << endl;
  }
  return 0;
}
