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 long long N = 1011;
const long long INF = 5e16;
namespace IO {
inline long long read() {
  long long x = 0;
  bool f = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar()) f ^= (ch == '-');
  for (; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ 48);
  return f ? -x : x;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 | 48);
}
inline void wrn(long long x) {
  write(x);
  putchar(' ');
}
inline void wln(long long x) {
  write(x);
  putchar('\n');
}
inline void wlnn(long long x, long long y) {
  wrn(x);
  wln(y);
}
}  // namespace IO
using namespace IO;
namespace Cesare {
long long x, k, p;
double Ans = 0, f[N][N];
int main() {
  x = read(), k = read(), p = read();
  double Mul = p * 0.01, Add = 1.0 - Mul;
  for (register int i = 0; i <= k; ++i)
    for (register int j = x + i; !(j & 1); j >>= 1) ++f[0][i];
  for (register int i = (1); i <= (k); ++i)
    for (register int j = (0); j <= (k); ++j) {
      if (j) f[i][j - 1] += f[i - 1][j] * Add;
      if (k >= (j << 1)) f[i][j << 1] += (f[i - 1][j] + 1) * Mul;
    }
  return printf("%.12lf\n", f[k][0]), 0;
}
}  // namespace Cesare
int main() { return Cesare::main(); }