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 Pascal (FPC 2.6.2) (judge verdict: Runtime Error)

uses math;
var
  n,i,j,s,t:longint;
  a:array[1..100]of longint;
begin
  readln(n);
  for i:=1 to n*2 do read(a[i]);
  for i:=1 to n*2-1 do
    for j:=i+1 to n*2 do
    if a[i]<a[j] then
    begin
      t:=a[i];
      a[i]:=a[j];
      a[j]:=t;
    end;
  for i:=1 to n do s:=s+min(a[i*2-1],a[i*2]);
  writeln(s);
end.