Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double eps(1e-8);
int x, k, p;
double dp[2][210];
int main() {
  scanf("%d %d %d", &x, &k, &p);
  double P = p / 100.;
  for (int i = 0; i <= k; i++) {
    int tmp = x + i;
    while (!(tmp & 1)) tmp >>= 1, dp[0][i] += 1.;
  }
  int now = 0;
  for (int i = 0; i < k; i++) {
    memset(dp[now ^ 1], 0, sizeof(dp[now ^ 1]));
    for (int j = 0; j <= k; j++) {
      dp[now ^ 1][j << 1] += (dp[now][j] + 1) * P;
      dp[now ^ 1][j] += dp[now][j + 1] * (1 - P);
    }
    now ^= 1;
  }
  printf("%.10f\n", dp[now][0]);
  return 0;
}