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;
int x, k, q, w, s;
double p;
double dp[205][265][2][265];
double f(int x, int mask, bool w, int cnt) {
  if (x == 0) {
    if (mask)
      for (int i = 0; i < 8; i++)
        if (mask & (1 << i)) return i;
    if (w) return 8;
    return 8 + cnt;
  }
  double &r = dp[x][mask][w][cnt];
  if (r >= 0) return r;
  r = 0;
  if (mask + 1 < 256)
    r += f(x - 1, mask + 1, w, cnt) * (1 - p);
  else {
    if (!w)
      r += f(x - 1, 0, 1, 1) * (1 - p);
    else
      r += f(x - 1, 0, 0, cnt) * (1 - p);
  }
  if (mask & (1 << 7)) {
    if (w)
      r += f(x - 1, mask * 2 % 256, 1, cnt + 1) * p;
    else
      r += f(x - 1, mask * 2 % 256, 1, 1) * p;
  } else {
    if (!w)
      r += f(x - 1, mask * 2, 0, cnt + 1) * p;
    else
      r += f(x - 1, mask * 2, 0, 1) * p;
  }
  return r;
}
int main() {
  scanf("%d %d %lf", &x, &k, &p);
  for (int i = 0; i < 205; i++)
    for (int j = 0; j < 265; j++)
      for (int k = 0; k < 2; k++)
        for (int l = 0; l < 265; l++) dp[i][j][k][l] = -1;
  p /= 100;
  w = ((x & (1 << 8)) ? 1 : 0);
  int i = 9;
  while (i <= 200 and ((x & (1 << i)) ? 1 : 0) == w) i++;
  printf("%.8lf", f(k, x % 256, w, i - 8));
}
