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;
const int INF = 1e9 + 10;
const long long int INFLL = 1e18 + 10;
const long double EPS = 1e-8;
const long double EPSLD = 1e-14;
const long long int MOD = 1000000007;
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
const int MAX_N = 256;
int x, n, ip;
long double p;
int a[MAX_N];
long double dp[2][MAX_N][MAX_N];
long double dp2[2][MAX_N];
long double ans;
int main() {
  scanf("%d %d %d", &x, &n, &ip);
  p = (long double)ip / 100.0;
  int nx = x;
  int it = 0;
  while (x > 0) {
    a[it++] = x % 2;
    x /= 2;
  }
  if (nx % 256 == 0) {
    for (int i = (8); i < (int)(MAX_N); i++) {
      if (a[i] == 1) {
        dp2[0][i - 8] = 1.0;
        break;
      }
    }
  } else {
    for (int i = (8); i < (int)(MAX_N); i++) {
      if (a[i] == 0) {
        dp[0][i - 8][nx % 256] = 1.0;
        break;
      }
    }
  }
  for (int i = (0); i < (int)(n); i++) {
    for (int j = (0); j < (int)(MAX_N); j++)
      for (int k = (0); k < (int)(MAX_N); k++) dp[(i + 1) % 2][j][k] = 0.0;
    for (int j = (0); j < (int)(MAX_N); j++) dp2[(i + 1) % 2][j] = 0.0;
    for (int j = (0); j < (int)(MAX_N); j++)
      for (int k = (0); k < (int)(MAX_N); k++)
        if (dp[i % 2][j][k] > EPSLD) {
          int nk = k + 1;
          if (nk == 256) {
            dp2[(i + 1) % 2][j] += dp[i % 2][j][k] * (1.0 - p);
          } else {
            dp[(i + 1) % 2][j][nk] += dp[i % 2][j][k] * (1.0 - p);
          }
          if (k / 128 == 1) {
            if (k % 128 == 0) {
              dp2[(i + 1) % 2][0] += dp[i % 2][j][k] * p;
            } else {
              dp[(i + 1) % 2][j + 1][(k * 2) % 256] += dp[i % 2][j][k] * p;
            }
          } else {
            dp[(i + 1) % 2][0][(k * 2) % 256] += dp[i % 2][j][k] * p;
          }
        }
    for (int j = (0); j < (int)(MAX_N); j++)
      if (dp2[i % 2][j] > EPSLD) {
        dp[(i + 1) % 2][0][1] += dp2[i % 2][j] * (1.0 - p);
        dp2[(i + 1) % 2][j + 1] += dp2[i % 2][j] * p;
      }
  }
  for (int j = (0); j < (int)(MAX_N); j++)
    for (int k = (0); k < (int)(MAX_N); k++)
      if (dp[n % 2][j][k] > EPSLD) {
        int nk = k;
        int res = 0;
        while (nk % 2 == 0) {
          res++;
          nk /= 2;
        }
        ans += dp[n % 2][j][k] * res;
      }
  for (int j = (0); j < (int)(MAX_N); j++)
    if (dp2[n % 2][j] > EPSLD) {
      ans += dp2[n % 2][j] * (8 + j);
    }
  printf("%.18lf\n", (double)ans);
  return 0;
}