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;
double dp[210][256][251][2];
int X, K, P;
double p, q;
int main() {
  scanf("%d%d%d", &X, &K, &P);
  p = double(P) / 100.0;
  q = 1 - p;
  memset(dp, 0, sizeof(dp));
  if (X < 256)
    dp[0][X][0][0] = 1;
  else {
    int x = X / 256, i;
    if (x % 2 == 0) {
      for (i = 0; x % 2 == 0; i++, x /= 2)
        ;
      dp[0][X % 256][i][0] = 1;
    } else {
      for (i = 0; x % 2 == 1; i++, x /= 2)
        ;
      dp[0][X % 256][i][1] = 1;
    }
  }
  for (int i = 0; i < K; i++) {
    for (int j = 0; j <= 255; j++)
      for (int l = 0; l <= 250; l++)
        for (int k = 0; k < 2; k++) {
          if (j & (1 << 7)) {
            if (k)
              dp[i + 1][(j * 2) % 256][l + 1][k] += p * dp[i][j][l][k];
            else
              dp[i + 1][(j * 2) % 256][1][1] += p * dp[i][j][l][k];
          } else {
            if (k)
              dp[i + 1][(j * 2) % 256][1][0] += p * dp[i][j][l][k];
            else if (l)
              dp[i + 1][(j * 2) % 256][l + 1][0] += p * dp[i][j][l][k];
            else
              dp[i + 1][(j * 2) % 256][0][0] = p * dp[i][j][l][k];
          }
          if (j < 255)
            dp[i + 1][j + 1][l][k] += q * dp[i][j][l][k];
          else if (k)
            dp[i + 1][0][l][0] += q * dp[i][j][l][k];
          else
            dp[i + 1][0][1][1] += q * dp[i][j][l][k];
        }
  }
  double ans = 0;
  for (int i = 0; i <= 255; i++)
    for (int j = 0; j <= 250; j++)
      for (int k = 0; k < 2; k++) {
        int num = 0;
        int temp = i, t = 0;
        while (temp % 2 == 0 && t++ <= 7) {
          num++;
          temp /= 2;
        }
        if (temp == 0 && k == 0) num += j;
        ans += num * dp[K][i][j][k];
      }
  printf("%.13lf\n", ans);
  return 0;
}