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>
double dp[201][501][256];
int cal(int i, int j) {
  int ans = 0;
  while (i % 2 == 0) {
    ans++;
    i /= 2;
    if (ans == 8) break;
  }
  if (ans < 8) return ans;
  if (j > 250) return ans + j - 250;
  return ans;
}
int main() {
  int x, k;
  double p;
  scanf("%d %d %lf", &x, &k, &p);
  p = p / 100.;
  int loc = x & 255;
  int base;
  if ((x & (1 << 8)) > 0) {
    base = 1;
    for (int i = 9;; i++) {
      if ((x & (1 << i)) == 0) break;
      base++;
    }
    dp[0][base][loc] = 1.;
  } else {
    if (x >= (1 << 8)) {
      base = 250;
      for (int i = 8;; i++) {
        if ((x & (1 << i)) > 0) break;
        base++;
      }
      dp[0][base][loc] = 1.;
    } else {
      dp[0][0][loc] = 1.;
    }
  }
  for (int step = 0; step < k; step++) {
    for (int i = 0; i <= 500; i++) {
      for (int j = 0; j < 256; j++) {
        if (dp[step][i][j] == 0) continue;
        if (i == 0) {
          if ((j & 128) > 0) {
            dp[step + 1][1][(j << 1) & 255] += dp[step][0][j] * p;
          } else {
            dp[step + 1][0][j << 1] += dp[step][0][j] * p;
          }
          if (j < 255) {
            dp[step + 1][0][j + 1] += dp[step][0][j] * (1 - p);
          } else {
            dp[step + 1][1][0] += dp[step][0][j] * (1 - p);
          }
          continue;
        }
        if (i <= 250) {
          if ((j & 128) > 0) {
            dp[step + 1][i + 1][(j << 1) & 255] += dp[step][i][j] * p;
          } else {
            dp[step + 1][251][(j << 1) & 255] += dp[step][i][j] * p;
          }
          if (j < 255) {
            dp[step + 1][i][j + 1] += dp[step][i][j] * (1 - p);
          } else {
            dp[step + 1][250 + i][0] += dp[step][i][j] * (1 - p);
          }
        } else {
          if ((j & 128) > 0) {
            dp[step + 1][1][(j << 1) & 255] += dp[step][i][j] * p;
          } else {
            dp[step + 1][i + 1][(j << 1) & 255] += dp[step][i][j] * p;
          }
          if (j < 255) {
            dp[step + 1][i][j + 1] += dp[step][i][j] * (1 - p);
          } else {
            dp[step + 1][1][0] += dp[step][i][j] * (1 - p);
          }
        }
      }
    }
  }
  double ans = 0.;
  for (int i = 0; i <= 500; i++) {
    for (int j = 0; j < 256; j++) {
      if (dp[k][i][j] == 0) continue;
      ans += dp[k][i][j] * cal(j, i);
    }
  }
  printf("%.13lf\n", ans);
  return 0;
}
