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: Wrong Answer)

#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

const int N=4010,P=1e9+7;

int n,a[N],b[N],fac[N],inv[N];
int f[N][N];

inline int &F(int x,int y){ return f[x+2005][y+2005]; }

inline void Pre(){
  fac[0]=1; for(int i=1;i<=4000;i++) fac[i]=1LL*fac[i-1]*i%P;
  inv[1]=1; for(int i=2;i<=4000;i++) inv[i]=1LL*(P-P/i)*inv[P%i]%P;
  inv[0]=1; for(int i=1;i<=4000;i++) inv[i]=1LL*inv[i]*inv[i-1]%P;
}

inline int C(int x,int y){
  return 1LL*fac[x]*inv[y]%P*inv[x-y]%P;
}

int main(){
  scanf("%d",&n); Pre();
  for(int i=1;i<=n;i++)
    scanf("%d%d",&a[i],&b[i]),F(-a[i],-b[i])++;
  for(int i=-2000;i<=2000;i++)
    for(int j=-2000;j<=2000;j++)
      F(i,j)=((long long)F(i,j)+F(i-1,j)+F(i,j-1))%P;
  int ans=0;
  for(int i=1;i<=n;i++) ans=(ans+F(a[i],b[i])-C(a[i]+a[i]+b[i]+b[i],b[i]+b[i]))%P;
  ans=1LL*ans*(P+1>>1)%P;
  printf("%d\n",(ans+P)%P);
  return 0;
}