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>
double f[2][256][2][250];
int n, st, p;
double prob;
int main() {
  scanf("%d%d%d", &st, &n, &p);
  prob = p / 100.0;
  if (st < 256)
    f[0][st][0][0] = 1.0;
  else {
    int cnt = 1;
    for (int i = 9; ((st >> i) & 1) == ((st >> 8) & 1); i++, cnt++)
      ;
    f[0][st & 255][(st >> 8) & 1][cnt] = 1.0;
  }
  int cur = 0;
  for (int i = 0; i < n; i++) {
    memset(f[cur ^ 1], 0, sizeof(f[cur ^ 1]));
    for (int j = 0; j < 256; j++) {
      for (int k = 0; k <= 240; k++) {
        if ((j >> 7) & 1)
          f[cur ^ 1][(j << 1) & 255][1][k + 1] += f[cur][j][1][k] * prob;
        else
          f[cur ^ 1][(j << 1) & 255][0][1] += f[cur][j][1][k] * prob;
        if (j == 255)
          f[cur ^ 1][0][0][k] += f[cur][j][1][k] * (1.0 - prob);
        else
          f[cur ^ 1][j + 1][1][k] += f[cur][j][1][k] * (1.0 - prob);
        if ((j >> 7) & 1)
          f[cur ^ 1][(j << 1) & 255][1][1] += f[cur][j][0][k] * prob;
        else
          f[cur ^ 1][(j << 1) & 255][0][k + 1] += f[cur][j][0][k] * prob;
        if (j == 255)
          f[cur ^ 1][0][1][1] += f[cur][j][0][k] * (1.0 - prob);
        else
          f[cur ^ 1][j + 1][0][k] += f[cur][j][0][k] * (1.0 - prob);
      }
    }
    cur ^= 1;
  }
  double ans = 0.0;
  for (int i = 0; i < 256; i++) {
    int s = 0;
    while (!((i >> s) & 1) && s < 8) ++s;
    for (int j = 0; j <= 240; j++)
      for (int k = 0; k < 2; k++) {
        int ss;
        if (k == 0 && s == 8)
          ss = s + j;
        else
          ss = s;
        ans += ss * f[cur][i][k][j];
      }
  }
  printf("%.10f\n", ans);
  return 0;
}