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;
const int MAXN = 300;
int x, n, p;
double dp[2][2][1 << 8][MAXN];
int main() {
  if (fopen("input.txt", "r")) freopen("input.txt", "r", stdin);
  cin >> x >> n >> p;
  int cnt = 0;
  for (int i = 9; 1; i++) {
    if (bool(x & 1 << i) == bool(x & 1 << 8))
      cnt++;
    else
      break;
  }
  dp[0][bool(x & 1 << 8)][x & (1 << 8) - 1][cnt] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 1 << 8; k++)
        for (int l = 0; l < MAXN; l++)
          if (dp[0][j][k][l] > 0) {
            double val = dp[0][j][k][l];
            if (k != (1 << 8) - 1)
              dp[1][j][k + 1][l] += val * (100. - p) / 100;
            else if (!j)
              dp[1][1][0][0] += val * (100. - p) / 100;
            else
              dp[1][0][0][l] += val * (100. - p) / 100;
            int b = bool(k & 1 << 7);
            if (b == j)
              dp[1][b][(k << 1) & (1 << 8) - 1][l + 1] += val * p / 100.;
            else
              dp[1][b][(k << 1) & (1 << 8) - 1][0] += val * p / 100.;
          }
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 1 << 8; k++)
        for (int l = 0; l < MAXN; l++)
          dp[0][j][k][l] = dp[1][j][k][l], dp[1][j][k][l] = 0;
  }
  double ans = 0;
  for (int j = 0; j < 2; j++)
    for (int k = 0; k < 1 << 8; k++)
      for (int l = 0; l < MAXN; l++)
        if (dp[0][j][k][l] > 0) {
          if (k == 0) {
            cnt = 8;
            if (!j) cnt += l + 1;
          } else
            cnt = __builtin_ctz(k);
          ans += cnt * dp[0][j][k][l];
        }
  cout << setprecision(10) << ans << '\n';
  return 0;
}
