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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
int n,m,ans,l[210];
inline int read(int &xx){
	xx=0;
	int f=0;char ch=getchar();
	for(;!isdigit(ch);f|=ch=='-',ch=getchar());
	for(;isdigit(ch);xx=(xx<<1)+(xx<<3)+ch-'0',ch=getchar());
	xx=f?-xx:xx;
}
void print1(int xx){
	if(xx>=10)print1(xx/10);
	putchar(xx%10+'0'); 
}
inline void print(int xx){
	if(xx<0){
		putchar('-');
		xx=-xx;
	}
	print1(xx);
	putchar('\n');
}
int main(){
	read(n);
	read(m);
	for(int i=1;i<=2*n;++i)cin>>l[i];
	sort(l+1,l+2*n+1);
	for(int i=1;i<=n;++i)ans+=l[2*i-1];
	print(ans);
	return 0;
}