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;
int n, ori;
double prob, dp[2][512][240];
int Count(int x) {
  int sum = 0;
  for (; ~x & 1; x >>= 1, ++sum)
    ;
  return sum;
}
double Dp() {
  double ans = 0;
  int cur = 0;
  prob /= 100;
  int x = ori >> 8, last = x & 1, s = 1;
  for (x >>= 1; x && (x & 1) == last; x >>= 1, ++s)
    ;
  dp[cur][ori & 511][s] = 1;
  for (int i = 1; i <= n + 1; ++i, cur ^= 1) {
    memset(dp[cur ^ 1], 0, sizeof(dp[cur ^ 1]));
    for (int j = 0; j <= 511; ++j)
      for (int k = 1; k <= i + 30; ++k) {
        int x = j >> 8 & 1;
        double &val = dp[cur][j][k];
        if (i == n + 1) {
          if (j == 0)
            ans += val * (8 + k);
          else
            ans += val * Count(j);
          continue;
        }
        int newj, newk, newx;
        newj = j * 2, newx = newj >> 8, newj %= 512;
        newk = newx % 2 == x ? k + 1 : 1;
        dp[cur ^ 1][newj][newk] += val * prob;
        newj = j + 1, newx = x, newk = k;
        if (newj == 512) newj = 0;
        dp[cur ^ 1][newj][newk] += val * (1 - prob);
      }
  }
  return ans;
}
int main() {
  cin >> ori >> n >> prob;
  printf("%.10f\n", Dp());
  return 0;
}