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
#include <bits/stdc++.h>
using namespace std;
int x, k, p, car[256];
long double dp[2][256][2][250];
const long double eps = 1e-12;
int main() {
  cin >> x >> k >> p;
  int last = 0, pp = 8;
  if (x > 255) {
    while (((x >> pp) & 1) == ((x >> 8) & 1)) last++, pp++;
    dp[0][x & 255][(x >> 8) & 1][last] = 1;
  } else
    dp[0][x & 255][0][1] = 1;
  long double p1 = p * 1.0 / 100, p2 = (100 - p) * 1.0 / 100;
  for (int i = 0; i < k; i++) {
    int now = i & 1, nxt = now ^ 1;
    memset(dp[nxt], 0, sizeof(dp[nxt]));
    for (int mask = 0; mask < 256; mask++) {
      for (int last = 0; last < 2; last++) {
        for (int cnt = 0; cnt < 250; cnt++) {
          if (dp[now][mask][last][cnt] > eps) {
            if ((mask & 128) == 0) {
              if (last == 0) {
                dp[nxt][mask << 1][0][cnt + 1] += dp[now][mask][last][cnt] * p1;
              } else {
                dp[nxt][mask << 1][0][1] += dp[now][mask][last][cnt] * p1;
              }
            } else {
              if (last == 0) {
                dp[nxt][(mask << 1) - 256][1][1] +=
                    dp[now][mask][last][cnt] * p1;
              } else {
                dp[nxt][(mask << 1) - 256][1][cnt + 1] +=
                    dp[now][mask][last][cnt] * p1;
              }
            }
            if (mask != 255) {
              dp[nxt][mask + 1][last][cnt] += dp[now][mask][last][cnt] * p2;
            } else {
              if (last == 0) {
                dp[nxt][0][1][1] += dp[now][mask][last][cnt] * p2;
              } else {
                dp[nxt][0][0][cnt] += dp[now][mask][last][cnt] * p2;
              }
            }
          }
        }
      }
    }
  }
  long double ans = 0;
  for (int i = 1; i < 256; i++) {
    int j = i;
    while (j % 2 == 0) car[i]++, j >>= 1;
  }
  int now = k & 1;
  for (int mask = 0; mask < 256; mask++) {
    for (int last = 0; last < 2; last++) {
      for (int cnt = 0; cnt < 250; cnt++) {
        if (dp[now][mask][last][cnt] > eps) {
          if (mask == 0) {
            ans += dp[now][mask][last][cnt] * (8 + (last == 0 ? cnt : 0));
          } else {
            ans += dp[now][mask][last][cnt] * car[mask];
          }
        }
      }
    }
  }
  cout << fixed << setprecision(13) << ans << endl;
  return 0;
}
