A. Little Pony and Expected Maximum

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, the m-th
face contains mdots. 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 dice n times.

Input

A single line contains two integers m and n (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 exceed 10  - 4.

Sample test(s)

input

6 1

output

3.500000000000

input

6 3

output

4.958333333333

input

2 2

output

1.750000000000

Note

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:

这里先推出每个数是当前序列的最大值的概率，一直推不出来，后来看了别人的思路懂了，最大值不超过i的所有序列总数是i^n，但是其中含有最大值不是i的序列，比如都是1，所以要减去最大值不超过i-1的所有序列的总数，最后结果是i^n-(i-1)^n;然后因为数字太大，所以每个都先除以m^n后再算。

#include<stdio.h>
#include<math.h>
int main()
{
int n,m;double sum;
scanf("%d%d",&m,&n);
sum=m;
for(int i=1;i<m;i++)
{
sum-=pow((double)1.0-(double)i/m,(double)n);
}
printf("%.12lf\n",sum);
return 0;
}