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 f[2001];
int main() {
  cin >> s >> p;
  for (int i = 0; i < s.length(); i++) {
    int j = i;
    int t = p.length() - 1;
    while (t >= 0 && j >= 0) {
      if (s[j] == p[t]) {
        j--;
        t--;
      } else {
        j--;
      }
    }
    if (t < 0) {
      f[i] = j + 1;
    } else {
      f[i] = -1;
    }
  }
  if (p.length() == 1 && s[0] == p[0]) {
    dp[0][0] = 1;
  }
  for (int i = 1; i < s.length(); i++) {
    dp[i][0] = dp[i - 1][0];
    if (f[i] >= 0 && i - f[i] + 1 == p.length()) {
      if (f[i] > 0) {
        dp[i][0] = max(dp[i][0], dp[f[i] - 1][0] + 1);
      } else {
        dp[i][0] = 1;
      }
    }
    for (int j = 1; j <= i; j++) {
      dp[i][j] = max(0, dp[i - 1][j]);
      if (f[i] >= 0) {
        int remove = i - f[i] + 1 - p.length();
        if (j >= remove) {
          if (f[i] > 0 && j - remove <= f[i]) {
            dp[i][j] = max(dp[i][j], dp[f[i] - 1][j - remove] + 1);
          } else {
            if (j == remove) {
              dp[i][j] = 1;
            }
          }
        }
      }
    }
  }
  for (int i = 0; i < s.length(); i++) {
    cout << dp[s.length() - 1][i] << " ";
  }
  cout << 0 << endl;
  return 0;
}
