CodeForces 453A

Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.

The dice has m faces: the first face of the dice contains a dot, the second one contains two dots, and so on, them-th face containsm dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability. Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dicen times.

Input

A single line contains two integers m andn (1 ≤ m, n ≤ 10⁵).

Output

Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed10 ^- 4.

Sample Input

Input

6 1

Output

3.500000000000

Input

6 3

Output

4.958333333333

Input

2 2

Output

1.750000000000

Hint

Consider the third test example. If you've made two tosses:

1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2.
2. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1.
3. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2.
4. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.

The probability of each outcome is 0.25, that is expectation equals to:

题目大意：

求出最大点数的期望（这两个字被平均代替后瞬间没有高尚感- -）。

思路：

若有5个的6面骰子出现最大点为1的次数为1^5,出现最大点为2的的次数为2^5-1^5,以此来推出最大点出现的次数为i^5-(i-1)^5;

求出最大值得期望，由示例可以得到 p = ( (1^n-0^n)*1 + (2^n-1^n)*2 。。。 +(m^n - (m-1)^n)*m ) / m^n ;将m^n带入 最终得到 p = m - ((m-1)/m)^n + ((m-2)/m)^n。。。+(1/m)^n ;

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<map>
#include<queue>
#include<algorithm>
#define LL long long
int a[1010];
using namespace std;
int main()
{
    LL s;
    int i,j,n,m,k,inr,x;
    double sum;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        sum=0;
        for(i=1;i<=m;i++)
        {
            sum+=i*(pow(i*1.0/m,n)-pow((i-1)*1.0/m,n));<span id="transmark"></span>
        }
        printf("%.12lf\n",sum);
    }
}