CodeForces

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

题意：给定X，现在进行K轮操作，每一轮有P%的概率加倍，(100-P)%的概率加一，问K轮之后的X的因子2的次数。

思路：首先一个数X中2的因子个数=转化为二进制后末尾0的个数=__builtin_ctz（X）；题解给的四维DP比较麻烦，这里有一种比较难以想到，但是不难理解的二维DP。

           用dp[i][j]，表示X+j进行i轮操作后的因子2的幂。 那么答案就是dp[K][0]；

那么转移就是:

            dp[i][j]+=(dp[i-1][j+1])*(1.0-P);   即第一次操作为+1；
            dp[i][j<<1]+=(dp[i-1][j]+1)*P;  即第一次操作为*2；

虽然乘法的话第二维会加倍增长，但加法的话第二维会减小1，所以最小影响到dp[K][0]的最大第二维就是K，而不用维护所有的dp[][j]，即j<=K；

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std; //*2 P
double dp[210][410],P;
int main()
{
    int X,K;
    scanf("%d%d%lf",&X,&K,&P); P/=100.0;
    rep(i,0,K) dp[0][i]=__builtin_ctz(X+i);
    rep(i,1,K){
        rep(j,0,K){
            dp[i][j]+=(dp[i-1][j+1])*(1.0-P);
            dp[i][j<<1]+=(dp[i-1][j]+1)*P;
        }
    }
    printf("%.10lf
",dp[K][0]);
    return 0;
}