Valera is a coder. Recently he wrote a funny program. The pseudo code for this program is given below:
    
    
      
    //input: integers x, k, p  
    a = x;  
    for(step = 1; step <= k; step = step + 1){  
        rnd = [random integer from 1 to 100];  
        if(rnd <= p)  
            a = a * 2;  
        else  
            a = a + 1;  
    }  
      
    s = 0;  
      
    while(remainder after dividing a by 2 equals 0){  
        a = a / 2;  
        s = s + 1;  
    }  
      
    

Now Valera wonders: given the values x, k and p, what is the expected value of the resulting number s?

Input

The first line of the input contains three integers x, k, p (1 ≤ x ≤ 109; 1 ≤ k ≤ 200; 0 ≤ p ≤ 100).

Output

Print the required expected value. Your answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.

Examples

Input

1 1 50


Output

1.0000000000000


Input

5 3 0


Output

3.0000000000000


Input

5 3 25


Output

1.9218750000000

Note

If the concept of expected value is new to you, you can read about it by the link: 

http://en.wikipedia.org/wiki/Expected_value
#include <bits/stdc++.h>
using namespace std;
double X, P;
int K;
double memo[205][205];
double v2(int x) {
  double ret = 0;
  while (x % 2 == 0) {
    x /= 2;
    ret++;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> X >> K >> P;
  for (int k = 0; k <= K; k++) {
    for (int i = 0; i <= K; i++) {
      if (k == 0) {
        memo[k][i] = v2(X + i);
        continue;
      }
      double prob = 0;
      for (int j = 0; j < k; j++) {
        prob = pow((100 - P) / 100, j) * P / 100;
        if ((i + j) % 2 == 0) {
          memo[k][i] += prob * (memo[k - (j + 1)][(i + j) / 2] + 1);
        }
      }
      prob = pow((100 - P) / 100, k);
      memo[k][i] += prob * v2(X + k + i);
    }
  }
  cout << setprecision(20) << memo[K][0] << endl;
  return 0;
}
