Problem:
Score : 200 points

Problem Statement

A shop sells N kinds of fruits, Fruit 1, \ldots, N, at prices of p_1, \ldots, p_N yen per item, respectively. (Yen is the currency of Japan.)

Here, we will choose K kinds of fruits and buy one of each chosen kind. Find the minimum possible total price of those fruits.

Constraints

1 \leq K \leq N \leq 1000

1 \leq p_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
p_1 p_2 \ldots p_N

Output

Print an integer representing the minimum possible total price of fruits.

Sample Input 1

5 3
50 100 80 120 80

Sample Output 1

210

This shop sells Fruit 1, 2, 3, 4, and 5 for 50 yen, 100 yen, 80 yen, 120 yen, and 80 yen, respectively.

The minimum total price for three kinds of fruits is 50 + 80 + 80 = 210 yen when choosing Fruit 1, 3, and 5.

Sample Input 2

1 1
1000

Sample Output 2

1000

Solution in Pascal (FPC 3.0.4) (judge verdict: Wrong Answer)

program main;
var
    n, k, sum, i: integer;
    a : array[1..1000] of integer;


procedure qs (l, r : integer);
var
    p, i, j, temp : integer;
begin
    p := a[(l + r) div 2]; i := l; j := r;
    
    while (i < j) do begin
        while (a[j] > p) do
            j := j - 1;
        while (a[i] < p) do
            i := i + 1;
        
        if (i < j) then begin
            temp := a[i];
            a[i] := a[j];
            a[j] := temp;
            j := j - 1;
            i := i + 1;
        end;
    end;

    if (l - r > 1) then begin
        qs(l, j);
        qs(i, r);
    end;
end;


begin
    readln(n, k);

    for i := 1 to n do 
        read(a[i]);

    qs(1, n);

    sum := 0;
    for i := 1 to k do
        sum := sum + a[i];
    

    writeln(sum);
end.