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;
double dp[205][2][250][1 << 8];
int main(void) {
  int x, K, p;
  scanf("%d %d %d", &x, &K, &p);
  if (x & (1 << 8)) {
    int p = 0;
    while (x & (1LL << (9 + p))) p++;
    dp[0][1][p][x & ((1 << 8) - 1)] = 1;
  } else {
    if ((1 << 9) > x)
      dp[0][0][0][x & ((1 << 8) - 1)] = 1;
    else {
      int p = 0;
      while (!(x & (1LL << (9 + p)))) p++;
      dp[0][0][p + 1][x & ((1 << 8) - 1)] = 1;
    }
  }
  double p1 = 1.0 * p / 100, p2 = 1 - p1;
  for (int i = 0; i < K; i++)
    for (int o = 0; o <= 1; o++)
      for (int j = 0; j < 250; j++)
        for (int k = 0; k < (1 << 8); k++)
          for (int op = 0; op <= 1; op++) {
            if (op == 0) {
              int b8 = k & (1 << 7);
              int nk = (k << 1) & ((1 << 8) - 1);
              if (b8) {
                if (o)
                  dp[i + 1][1][j + 1][nk] += dp[i][o][j][k] * p1;
                else
                  dp[i + 1][1][0][nk] += dp[i][o][j][k] * p1;
              } else {
                if (o)
                  dp[i + 1][0][1][nk] += dp[i][o][j][k] * p1;
                else if (!j)
                  dp[i + 1][0][0][nk] += dp[i][o][j][k] * p1;
                else
                  dp[i + 1][0][j + 1][nk] += dp[i][o][j][k] * p1;
              }
            } else {
              int nk = (k + 1) & ((1 << 8) - 1);
              int b8 = k & (1 << 7), nb8 = nk & (1 << 7);
              if (b8 && !nb8) {
                if (o)
                  dp[i + 1][0][j + 1][nk] += dp[i][o][j][k] * p2;
                else
                  dp[i + 1][1][0][nk] += dp[i][o][j][k] * p2;
              } else
                dp[i + 1][o][j][nk] += dp[i][o][j][k] * p2;
            }
          }
  double ans = 0;
  for (int o = 0; o <= 1; o++)
    for (int j = 0; j < 250; j++)
      for (int k = 0; k < (1 << 8); k++) {
        if (!k) {
          if (!o)
            ans += dp[K][o][j][k] * (8 + j);
          else
            ans += dp[K][o][j][k] * 8;
        } else {
          int cnt = 0;
          while (!(k & (1 << cnt))) cnt++;
          ans += dp[K][o][j][k] * cnt;
        }
      }
  printf("%.10f\n", ans);
  return 0;
}