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: Wrong Answer)

var
  n,i,j,t,s,s1:longint;
  a:array[1..200] of longint;
begin
  readln(n);
  for i:=1 to n*2 do read(a[i]);
  for i:=1 to n-1 do
    for j:=i+1 to n do
      if a[i]>a[j] then
        begin
          t:=a[i]; a[i]:=a[j]; a[j]:=t;
        end;
  for i:=1 to n*2 do if odd(n) then inc(s,a[i]) else inc(s1,a[i]);
  if s<s1 then writeln(s) else writeln(s1);
end.