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;
long double dp[2][2][255][256];
int mm[256];
int main() {
  mm[0] = -1;
  for (int i = 1; i < 256; i++) mm[i] = mm[i - 1] + ((i & (i - 1)) == 0);
  int x, k, tp;
  cin >> x >> k >> tp;
  long double p = tp / 100.0;
  int now = 0, la = 1;
  if (~x >> 8 & 1) {
    int loc = 0;
    for (int i = 30; i >= 8; i--)
      if (x >> i & 1) loc = i - 8;
    dp[now][0][loc][x % 256] = 1;
  } else {
    int cnt = 0;
    for (int i = 30; i >= 8; i--) cnt = (x >> i & 1 ? cnt + 1 : 0);
    dp[now][1][cnt][x % 256] = 1;
  }
  for (int _ = 1; _ <= k; _++) {
    swap(now, la);
    memset(dp[now], 0, sizeof(dp[now]));
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 250; j++) {
        for (int k = 0; k < 256; k++) {
          if (i ^ (k < 128))
            dp[now][i][j + 1][k % 128 * 2] += dp[la][i][j][k] * p;
          else
            dp[now][i ^ 1][1][k % 128 * 2] += dp[la][i][j][k] * p;
        }
        for (int k = 0; k < 255; k++)
          dp[now][i][j][k + 1] += dp[la][i][j][k] * (1 - p);
        if (i == 0)
          dp[now][1][1][0] += dp[la][0][j][255] * (1 - p);
        else
          dp[now][0][j][0] += dp[la][1][j][255] * (1 - p);
      }
  }
  long double res = 0;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 250; j++)
      for (int k = 0; k < 256; k++)
        res += dp[now][i][j][k] * (k ? mm[k & -k] : 8 + (i == 0) * j);
  cout << fixed << setprecision(12) << res << endl;
  return 0;
}