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 int MaxN = 200;
const int MaxM = MaxN + 32;
const int M = 255;
int x, n, p;
double f[MaxN + 1][M + 1][MaxM + 1][2];
inline void initialize() {
  int up = x >> 8;
  int bit = up & 1, cnt = 0;
  while (up && (up & 1) == bit) up >>= 1, ++cnt;
  f[0][x & M][cnt][bit] = 1.0;
}
int main() {
  cin >> x >> n >> p;
  double p1 = p * 0.01;
  double p2 = 1.00 - p1;
  initialize();
  for (int i = 0; i < n; ++i)
    for (int s = 0; s <= M; ++s)
      for (int j = 0; j < MaxM; ++j)
        for (int k = 0; k < 2; ++k) {
          double now = f[i][s][j][k];
          int bit = s >> 7 & 1;
          int cnt = bit == k ? j + 1 : 1;
          f[i + 1][s << 1 & M][cnt][bit] += now * p1;
          bit = s == M ? !k : k;
          cnt = s == M && !k ? 1 : j;
          f[i + 1][s + 1 & M][cnt][bit] += now * p2;
        }
  double res = 0.0;
  for (int s = 0; s <= M; ++s)
    for (int i = 0; i < MaxM; ++i)
      for (int j = 0; j < 2; ++j)
        res += f[n][s][i][j] * (s ? __builtin_ctz(s) : 8 + (j ? 0 : i));
  printf("%.8f\n", res);
  return 0;
}