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[201][256][300][2];
int main() {
  int x, k, p, i, j, l, r;
  double ans = 0;
  scanf("%d %d %d", &x, &k, &p);
  if (x & (1 << 8)) {
    int y = x >> 8, c = 0;
    while (y) {
      if (y % 2 == 0) break;
      y >>= 1;
      c++;
    }
    dp[0][x % 256][c][1] = 1;
  } else {
    int y = x >> 8, c = 0;
    while (y) {
      if (y % 2 == 1) break;
      y >>= 1;
      c++;
    }
    dp[0][x % 256][c][0] = 1;
  }
  for (i = 0; i < k; i++) {
    for (j = 0; j < 256; j++) {
      for (l = 0; l < 300; l++) {
        for (r = 0; r < 2; r++) {
          double p1, p2;
          if (dp[i][j][l][r] == 0) continue;
          p1 = dp[i][j][l][r] * p / 100;
          p2 = dp[i][j][l][r] * (100 - p) / 100;
          if (j * 2 >= 256) {
            if (r == 0) {
              dp[i + 1][j * 2 % 256][1][1] += p1;
            } else {
              dp[i + 1][j * 2 % 256][l + 1][1] += p1;
            }
          } else {
            if (r == 0) {
              dp[i + 1][j * 2 % 256][l + 1][0] += p1;
            } else {
              dp[i + 1][j * 2 % 256][1][0] += p1;
            }
          }
          if (j + 1 == 256) {
            if (r == 0) {
              dp[i + 1][0][1][1] += p2;
            } else {
              dp[i + 1][0][l][0] += p2;
            }
          } else {
            dp[i + 1][j + 1][l][r] += p2;
          }
        }
      }
    }
  }
  for (i = 0; i < 256; i++) {
    for (j = 0; j < 300; j++) {
      for (l = 0; l < 2; l++) {
        if (i == 0) {
          ans += dp[k][i][j][l] * (j * (1 - l) + 8);
        } else {
          ans += dp[k][i][j][l] * __builtin_ctz(i);
        }
      }
    }
  }
  printf("%.12lf\n", ans);
  return 0;
}