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;
int x, n, p;
int a[1005];
long double f[205][205][205];
int check(int m) {
  int s = 0;
  while (m % 2 == 0) {
    s++;
    m /= 2;
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, k;
  long double xx, yy;
  cin >> x >> n >> p;
  xx = (long double)p / 100.0;
  yy = (long double)(100 - p) / 100.0;
  for (i = 0; i <= 1000; i++) a[i] = check(x + i);
  for (j = 0; j <= 201; j++) {
    for (k = 0; k <= 201; k++) {
      f[0][j][k] = (long double)(a[j] + k);
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= 200; j++) {
      for (k = 0; k <= 200; k++) {
        f[i][j][k] = f[i - 1][j + 1][k] * yy;
        if (j % 2 == 0)
          f[i][j][k] += (f[i - 1][j / 2][k + 1] * xx);
        else
          f[i][j][k] += ((long double)k * xx);
      }
    }
  }
  cout.precision(9);
  cout << fixed << f[n][0][0] << endl;
  return 0;
}