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 = (int)1e9;
const double INF = 1e12, EPS = 1e-9;
int x, r;
double p;
double dp[2][256][300][2];
int main() {
  cin >> x >> r >> p;
  p /= 100;
  int rem = x % 256;
  x /= 256;
  if (x & 1) {
    dp[0][rem][__builtin_ctz(x + 1)][1] = 1;
  } else {
    if (x) x = __builtin_ctz(x);
    dp[0][rem][x][0] = 1;
  }
  int cur = 0, next = 1;
  for (int i = 0; i < (int)r; i++) {
    memset(dp[next], 0, sizeof(dp[next]));
    for (int j = 0; j < (int)256; j++)
      for (int l = 0; l < (int)300; l++)
        for (int one = 0; one < (int)2; one++)
          if (dp[cur][j][l][one])
            for (int k = 0; k < (int)2; k++) {
              int nj = (k ? j * 2 : j + 1) % 256;
              int carry = (k ? j * 2 : j + 1) / 256;
              int none = one, nl = l;
              if (carry) {
                if (k) {
                  if (none)
                    nl++;
                  else
                    none = 1, nl = 1;
                } else {
                  if (none)
                    none = 0;
                  else
                    none = 1, nl = 1;
                }
              } else {
                if (k) {
                  if (none)
                    none = 0, nl = 1;
                  else if (nl)
                    nl++;
                }
              }
              dp[next][nj][nl][none] += (k ? p : 1 - p) * dp[cur][j][l][one];
            }
    swap(cur, next);
  }
  double ans = 0;
  for (int j = 0; j < (int)256; j++)
    for (int l = 0; l < (int)300; l++)
      for (int one = 0; one < (int)2; one++)
        if (dp[cur][j][l][one]) {
          double p = dp[cur][j][l][one];
          if (j)
            ans += p * __builtin_ctz(j);
          else {
            if (one) ans += p * 8;
            if (!one && l) ans += p * (8 + l);
          }
        }
  printf("%.20f\n", ans);
  return 0;
}