Problem:
Score : 200 points

Problem Statement

Snuke is having a barbeque party.

At the party, he will make N servings of Skewer Meal.

Example of a serving of Skewer Meal

He has a stock of 2N skewers, all of which will be used in Skewer Meal. The length of the i-th skewer is L_i.
Also, he has an infinite supply of ingredients.

To make a serving of Skewer Meal, he picks 2 skewers and threads ingredients onto those skewers.
Let the length of the shorter skewer be x, then the serving can hold the maximum of x ingredients.

What is the maximum total number of ingredients that his N servings of Skewer Meal can hold, if he uses the skewers optimally?

Constraints

1≦N≦100

1≦L_i≦100

For each i, L_i is an integer.

Input

The input is given from Standard Input in the following format:

N
L_1 L_2 ... L_{2N}

Output

Print the maximum total number of ingredients that Snuke's N servings of Skewer Meal can hold.

Sample Input 1

2
1 3 1 2

Sample Output 1

3

If he makes a serving using the first and third skewers, and another using the second and fourth skewers, each serving will hold 1 and 2 ingredients, for the total of 3.

Sample Input 2

5
100 1 2 3 14 15 58 58 58 29

Sample Output 2

135

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

const int MAXN = 100 + 5;

int N, Ans;
int A[MAXN << 1];

inline int read()
{
	register int x = 0;
	register char ch = getchar();
	while(!isdigit(ch)) ch = getchar();
	while(isdigit(ch))
	{
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x;
}

int main()
{
	N = read() << 1;
	for(int i = 1; i <= N; ++i) A[i] = read();
	std::sort(A + 1, A + 1 + N);
	for(int i = 1; i <= N; i += 2) Ans += A[i];
	printf("%lld\n", Ans);
	return 0;
}