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)

// eddy1021
#include <bits/stdc++.h>
using namespace std;
typedef double D;
typedef long double LD;
typedef long long ll;
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef pair<LD,LD> Pt;
typedef tuple<int,int,int> tiii;
typedef tuple<LL,LL,LL> tlll;
#define mod9 1000000009ll
#define mod7 1000000007ll
#define INF  1023456789ll
#define INF16 10000000000000000ll
#define FI first
#define SE second
#define X FI
#define Y SE
#define PB push_back
#define MP make_pair
#define MT make_tuple
#define eps 1e-9
#define SZ(x) (int)(x).size()
#define ALL(x) (x).begin(), (x).end()
#ifndef ONLINE_JUDGE
#define debug(...) printf(__VA_ARGS__)
#else 
#define debug(...)
#endif
inline ll getint(){
  ll _x=0,_tmp=1; char _tc=getchar();    
  while( (_tc<'0'||_tc>'9')&&_tc!='-' ) _tc=getchar();
  if( _tc == '-' ) _tc=getchar() , _tmp = -1;
  while(_tc>='0'&&_tc<='9') _x*=10,_x+=(_tc-'0'),_tc=getchar();
  return _x*_tmp;
}
inline ll add( ll _x , ll _y , ll _mod = mod7 ){
  ll _ = _x + _y;
  if( _ >= _mod ) _ -= _mod;
  return _;
}
inline ll sub( ll _x , ll _y , ll _mod = mod7 ){
  ll _ = _x - _y;
  if( _ < 0 ) _ += _mod;
  return _;
}
inline ll mul( ll _x , ll _y , ll _mod = mod7 ){
  ll _ = _x * _y;
  if( _ >= _mod ) _ %= _mod;
  return _;
}
ll mypow( ll _a , ll _x , ll _mod ){
  if( _x == 0 ) return 1ll;
  ll _tmp = mypow( _a , _x / 2 , _mod );
  _tmp = mul( _tmp , _tmp , _mod );
  if( _x & 1 ) _tmp = mul( _tmp , _a , _mod );
  return _tmp;
}
ll mymul( ll _a , ll _x , ll _mod ){
  if( _x == 0 ) return 0ll;
  ll _tmp = mymul( _a , _x / 2 , _mod );
  _tmp = add( _tmp , _tmp , _mod );
  if( _x & 1 ) _tmp = add( _tmp , _a , _mod );
  return _tmp;
}
inline bool equal( D _x ,  D _y ){
  return _x > _y - eps && _x < _y + eps;
}
int __ = 1 , _cs;
/*********default*********/
#define N 111
void build(){

}
int n;
vector<int> v;
void init(){
  n = getint();
  for( int i = 0 ; i < n + n ; i ++ )
    v.push_back( getint() );
  sort( ALL( v ) );
}
void solve(){
  int ans = 0;
  for( int i = 0 ; i < n + n ; i += 2 )
    ans += v[ i ];
  cout << ans << endl;
}
int main(){
  build();
  //__ = getint();
  while( __ -- ){
    init();
    solve();
  }
}