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 LLINF = 1e18;
const int INF = 1e9;
const double EPS = 1e-7;
inline const int &INT() {
  static int x;
  scanf("%d", &x);
  return x;
}
inline const long long &LONG() {
  static long long x;
  scanf("%lld", &x);
  return x;
}
template <class T>
inline void MIN(T &a, const T &b) {
  if (b < a) a = b;
}
template <class T>
inline void MAX(T &a, const T &b) {
  if (b > a) a = b;
}
const int NN = 203;
double F[2][NN];
int X, K, P;
double aa, bb;
int main() {
  cin >> X >> K >> P;
  aa = 1.0 * P / 100;
  bb = 1.0 * (100 - P) / 100;
  int cur = 0;
  for (int i = (0); i <= (K); ++i) F[cur][i] = 0;
  F[cur][0] = 1;
  double ans = 0;
  for (int zz = 0; zz < (K); ++zz) {
    F[cur ^ 1][0] = 0;
    for (int i = (0); i <= (K); ++i) F[cur ^ 1][i + 1] = F[cur][i] * bb;
    for (int i = 0; i <= K; i += 2) {
      F[cur ^ 1][i / 2] += F[cur][i] * aa;
      ans += F[cur][i] * aa;
    }
    cur ^= 1;
  }
  for (int i = (0); i <= (K); ++i) {
    int z = X + i, c = 0;
    while (z % 2 == 0) z /= 2, ++c;
    ans += F[cur][i] * c;
  }
  printf("%.13lf\n", ans);
  return 0;
}