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[2][2][250][256];
int get1(int n) {
  int ret = 0;
  while (n) {
    if (n & 1)
      ret++;
    else
      return ret;
    n >>= 1;
  }
  return ret;
}
int get0(int n) {
  if (n == 0) return 1;
  int ret = 0;
  while (n) {
    if (n & 1)
      return ret;
    else
      ret++;
    n >>= 1;
  }
  return ret;
}
int main() {
  int n, i, j, x, k;
  double p;
  scanf("%d%d%d", &x, &k, &i);
  p = i / 100.0;
  if (x & 256)
    dp[0][1][get1(x >> 8)][x & 255] = 1;
  else
    dp[0][0][get0(x >> 8)][x & 255] = 1;
  for (int ii = 0; ii < k; ii++) {
    int cur = (ii & 1);
    memset(dp[cur ^ 1], 0, sizeof(dp[cur ^ 1]));
    for (i = 1; i < 250; i++) {
      for (j = 0; j < 128; j++) {
        dp[cur ^ 1][0][i + 1][j << 1] += dp[cur][0][i][j] * p;
        dp[cur ^ 1][0][i][j + 1] += dp[cur][0][i][j] * (1 - p);
      }
      for (; j < 256; j++) {
        dp[cur ^ 1][1][1][(j & 127) << 1] += dp[cur][0][i][j] * p;
        if (j == 255)
          dp[cur ^ 1][1][1][0] += dp[cur][0][i][j] * (1 - p);
        else
          dp[cur ^ 1][0][i][j + 1] += dp[cur][0][i][j] * (1 - p);
      }
    }
    for (i = 1; i < 250; i++) {
      for (j = 0; j < 128; j++) {
        dp[cur ^ 1][0][1][j << 1] += dp[cur][1][i][j] * p;
        dp[cur ^ 1][1][i][j + 1] += dp[cur][1][i][j] * (1 - p);
      }
      for (; j < 256; j++) {
        dp[cur ^ 1][1][i + 1][(j & 127) << 1] += dp[cur][1][i][j] * p;
        if (j == 255)
          dp[cur ^ 1][0][i][0] += dp[cur][1][i][j] * (1 - p);
        else
          dp[cur ^ 1][1][i][j + 1] += dp[cur][1][i][j] * (1 - p);
      }
    }
  }
  int cur = (k & 1);
  double ans = 0;
  for (i = 1; i < 250; i++) {
    for (j = 0; j < 256; j++) {
      if (j == 0) {
        ans += (i + 8) * dp[cur][0][i][j];
        ans += 8 * dp[cur][1][i][j];
      } else {
        ans += get0(j) * dp[cur][0][i][j];
        ans += get0(j) * dp[cur][1][i][j];
      }
    }
  }
  printf("%.12lf\n", ans);
}