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
#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 9, M = 250;
double f[2][N][M];
int x, n, s0, k0;
double p;
void get0(int s, int k, int &s1, int &k1) {
  if (((s >> 7) & 1) == ((s >> 8) & 1))
    k++;
  else
    k = 1;
  (s <<= 1) %= N;
  s1 = s;
  k1 = k;
}
void get1(int s, int k, int &s1, int &k1) {
  if (s == (1 << 8) - 1) k = 1;
  (s += 1) %= N;
  s1 = s;
  k1 = k;
}
int main() {
  scanf("%d%d%lf", &x, &n, &p);
  p /= 100;
  s0 = x % N;
  if (x & (1 << 8)) {
    int down = 8;
    while (x & (1 << down)) k0++, down++;
  } else {
    int down = 8;
    while (down <= 30 && (x & (1 << down)) == 0) k0++, down++;
  }
  int pre = 0, down = 1;
  f[down][s0][k0] = 1;
  for (int st = 1; st <= n; ++st) {
    swap(pre, down);
    memset(f[down], 0, sizeof(f[down]));
    for (int s = 0; s < N; ++s)
      for (int k = 0; k <= M - 1; ++k)
        if (f[pre][s][k] > 0) {
          int s1, k1;
          get0(s, k, s1, k1);
          f[down][s1][k1] += p * f[pre][s][k];
          get1(s, k, s1, k1);
          f[down][s1][k1] += (1.0 - p) * f[pre][s][k];
        }
  }
  double ans = 0;
  for (int s = 0; s <= N - 1; ++s)
    for (int k = 0; k <= M - 1; ++k)
      if (s) {
        int i = 0;
        while ((s & (1 << i)) == 0) i++;
        ans += f[down][s][k] * i;
      } else
        ans += f[down][s][k] * (k + 8.0);
  printf("%.10lf\n", ans);
  return 0;
}
