Drinks Watermelon

B. Drinks

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are ndrinks in his fridge, the volume fraction of orange juice in the i-th drink equals pi percent.

One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the n drinks and mixed them. Then he wondered, how much orange juice the cocktail has.

Find the volume fraction of orange juice in the final drink.

Input

The first input line contains a single integer n (1 ≤ n ≤ 100) — the number of orange-containing drinks in Vasya's fridge. The second line contains n integers pi (0 ≤ pi ≤ 100) — the volume fraction of orange juice in the i-th drink, in percent. The numbers are separated by a space.

Output

Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10  - 4.

Sample test(s)

input

3
50 50 100

output

66.666666666667

input

4
0 25 50 75

output

37.500000000000

Note

Note to the first sample: let's assume that Vasya takes x milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal  milliliters. The total cocktail's volume equals 3·x milliliters, so the volume fraction of the juice in the cocktail equals , that is, 66.(6) percent.

#include<iostream>
#include<cstdio>
using namespace std;

int main(){
	int n;
	cin>>n;
	double d;
	double sum = 0.0;
	for(int i = 0; i < n; i++){
		cin>>d;
		sum+=d;
	}
	printf("%.12f\n",sum/n);
	return 0;
}

A. Watermelon

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed w kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.

Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.

Input

The first (and the only) input line contains integer number w (1 ≤ w ≤ 100) — the weight of the watermelon bought by the boys.

Output

Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.

Sample test(s)

input

8

output

YES

Note

For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).

#include<iostream>
using namespace std;

int main(){
	int n;
	cin>>n;
	if(n==2){
		cout<<"NO"<<endl;
		return 0;
	}
	if(n&1){
		cout<<"NO"<<endl;
	}else{
		cout<<"YES"<<endl;
	}
	return 0;
}