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 = 0x3fffffff;
const int MOD = 1000000007;
int x, k, p;
long double dp[2][1 << 8][256][2];
int main() {
  scanf("%d%d%d", &x, &k, &p);
  int pre = 1, now = 0, cnt = 0;
  if ((x >> 8) & 1)
    for (int i = 8; i < 31; i++)
      if ((1 << i) & x)
        cnt++;
      else
        break;
  else
    for (int i = 8; i < 31; i++)
      if ((1 << i) & x)
        break;
      else
        cnt++;
  ;
  int mask = (1 << 8) - 1;
  dp[pre][mask & x][cnt][(x >> 8) & 1] = 1.0;
  long double a = p / 100.0, b = (100 - p) / 100.0;
  for (int i = 0; i < k; i++) {
    memset(dp[now], 0, sizeof(dp[now]));
    for (int j = 0; j < (1 << 8); j++) {
      for (int o = 0; o < 256; o++) {
        for (int z = 0; z < 2; z++) {
          int _o = (((j >> 7) & 1) == z) ? o + 1 : 1;
          dp[now][(j << 1) & mask][_o][(j >> 7) & 1] += dp[pre][j][o][z] * a;
          int st = j + 1, _z = z;
          _o = o;
          if (((1 << 7) & j) && !((1 << 7) & st)) {
            _z = z ^ 1;
            if (_z == 1) _o = 1;
          } else
            _o = o;
          dp[now][st & mask][_o][_z] += dp[pre][j][o][z] * b;
        }
      }
    }
    swap(pre, now);
  }
  double ans = 0;
  for (int j = 0; j < (1 << 8); j++) {
    for (int o = 0; o < 256; o++) {
      for (int z = 0; z < 2; z++) {
        cnt = 0;
        for (int i = 0; i < 8; i++) {
          if (!((1 << i) & j))
            cnt++;
          else
            break;
        }
        if (cnt == 8 && z == 0) cnt += o;
        ans += dp[pre][j][o][z] * cnt;
      }
    }
  }
  printf("%.9lf\n", ans);
  return 0;
}