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;
namespace IO {
const int sz = 1 << 20;
char a[sz + 5], b[sz + 5], *p1 = a, *p2 = a, *t = b, p[105];
inline char gc() {
  return getchar();
  if (p1 == p2) p2 = (p1 = a) + fread(a, 1, sz, stdin);
  return *p1++;
}
template <class T>
inline T read() {
  T ret = 0;
  char c = gc();
  int f = 1;
  for (; c < '0' || c > '9'; c = gc())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = gc()) ret = ret * 10 + (c - '0');
  return ret * f;
}
template <class T>
inline void read(T& ret) {
  ret = read<T>();
}
inline void flush() { fwrite(b, 1, t - b, stdout), t = b; }
inline void pc(char x) {
  *t++ = x;
  if (t - b == sz) flush();
}
template <class T>
inline void print(T x, char c = '\n') {
  if (x < 0) x = -x, pc('-');
  if (x == 0) pc('0');
  int t = 0;
  for (; x; x /= 10) p[++t] = x % 10 + '0';
  for (; t; --t) pc(p[t]);
  pc(c);
}
struct F {
  ~F() { flush(); }
} f;
}  // namespace IO
using IO::print;
using IO::read;
int n, k;
double p;
double f[310][310];
int main() {
  read(n), read(k);
  p = (double)read<int>() / 100.0;
  for (int i = 0; i <= k; i++)
    for (int j = n + i; !(j & 1); j >>= 1) f[0][i]++;
  for (int i = 0; i < k; i++)
    for (int j = 0; j <= k; j++) {
      if (j) f[i + 1][j - 1] += (1.0 - p) * f[i][j];
      if (j * 2 <= k) f[i + 1][j * 2] += p * (f[i][j] + 1);
    }
  printf("%.6lf\n", f[k][0]);
  return 0;
}