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;
template <typename T>
void pv(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
template <typename T>
void chmin(T &t, const T &f) {
  if (t > f) t = f;
}
template <typename T>
void chmax(T &t, const T &f) {
  if (t < f) t = f;
}
int in() {
  int x;
  scanf("%d", &x);
  return x;
}
int X, K;
double P;
double dp[210][250][250];
int main() {
  for (; ~scanf("%d%d", &X, &K);) {
    P = in() / 100.0;
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1.0;
    double ans = 0.0;
    for (int k = 0; k < K; ++k)
      for (int p = 0; p < 35 + K; ++p)
        for (int s = 0; s < 35 + K; ++s) {
          if (s % 2 != 0) {
            ans += (dp[k][p][s] * P) * p;
          } else {
            dp[k + 1][p + 1][s >> 1] += dp[k][p][s] * P;
          }
          dp[k + 1][p][s + 1] += dp[k][p][s] * (1.0 - P);
        }
    for (int p = 0; p < 35 + K; ++p)
      for (int s = 0; s < 35 + K; ++s) {
        ans += dp[K][p][s] * (p + __builtin_ctz(X + s));
      }
    printf("%.10f\n", ans);
  }
  return 0;
}