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 int N = 202;
double dp[N][1 << 8][2][255];
bool vis[N][1 << 8][2][255];
int x, n, mask = (1 << 8) - 1;
double p, q;
double dfs(int i, int j, int k, int t) {
  if (vis[i][j][k][t]) return dp[i][j][k][t];
  vis[i][j][k][t] = 1;
  if (i == n) {
    if (j == 0) {
      dp[i][j][k][t] = 8 + (k == 0 ? t : 0);
    } else {
      for (int a = 0; (j & 1 << a) == 0; ++a) ++dp[i][j][k][t];
    }
    return dp[i][j][k][t];
  }
  if (j == mask) {
    int nk, nt;
    if (k == 0) {
      nk = 1;
      nt = 1;
    } else {
      nk = 0;
      nt = t;
    }
    dp[i][j][k][t] += q * dfs(i + 1, 0, nk, nt);
  } else {
    dp[i][j][k][t] += q * dfs(i + 1, j + 1, k, t);
  }
  int nk = (j & 1 << 7) != 0, nt;
  if (nk == k) {
    if (t == 0)
      nt = 0;
    else
      nt = t + 1;
  } else
    nt = 1;
  return dp[i][j][k][t] += p * dfs(i + 1, (j << 1) & mask, nk, nt);
}
int main() {
  cin >> x >> n >> p;
  p /= 100;
  q = 1 - p;
  int i = 0, j = x & mask, k = (x & (1 << 8)) != 0, t = 0;
  if (x > mask) {
    for (int a = 8; ((x & 1 << a) != 0) == k; ++a) ++t;
  }
  printf("%.8f\n", dfs(i, j, k, t));
  return 0;
}