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, K, x, p, q;
long double dp[2][256][2][225], ans;
int cal(int x) {
  int res = 0;
  while (x % 2 == 0) {
    res++;
    x /= 2;
  }
  return res;
}
int main() {
  cin >> x >> K >> p;
  q = 100 - p;
  int t = 0, y = x >> 8 & 1, xx = x;
  x /= 256;
  if (x == 0) t = 1;
  while (x) {
    if ((x & 1) == y)
      t++;
    else
      break;
    x /= 2;
  }
  x = xx;
  dp[0][x & 255][x >> 8 & 1][t] = 1;
  for (int i = 0; i < K; i++) {
    int now = i & 1, nxt = i + 1 & 1;
    memset(dp[nxt], 0, sizeof(dp[nxt]));
    for (int j = 0; j <= 255; j++) {
      for (int k = 1; k < 225; k++) {
        for (int l = 0; l < 2; l++) {
          if ((j >> 7 & 1) == l)
            t = k + 1;
          else
            t = 1;
          dp[nxt][(j << 1) & 255][j >> 7 & 1][t] +=
              dp[now][j][l][k] * p / 100.0;
          if (j + 1 > 255) {
            if (l == 0) {
              dp[nxt][0][1][1] += dp[now][j][l][k] * q / 100.0;
            } else {
              dp[nxt][0][0][k] += dp[now][j][l][k] * q / 100.0;
            }
          } else {
            dp[nxt][j + 1][l][k] += dp[now][j][l][k] * q / 100.0;
          }
        }
      }
    }
  }
  K = K & 1;
  for (int i = 0; i <= 255; i++) {
    for (int j = 1; j < 225; j++) {
      for (int k = 0; k < 2; k++) {
        if (i == 0) {
          if (k == 0) {
            ans += (8 + j) * dp[K][i][k][j];
          } else {
            ans += 8 * dp[K][i][k][j];
          }
        } else {
          ans += dp[K][i][k][j] * cal(i);
        }
      }
    }
  }
  printf("%.10lf\n", (double)(ans));
}