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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
int main(void) {
    
    int k, n;
    scanf("%d%d\n", &n, &k);

    if ((k < 1) || (1000 < k) || (n < 1) || (1000 < n)){return -1;}

    int p[n], i=0, j, temp;
    while (i<n){
        scanf("%d", &p[i]);
        if ((p[i] < 1) || (1000 < p[i])){return -1;}  
        i++;
    }

    int sum = 0;

    i = 0;
    while (i < k)
    {
        j = i+1;
        while (j<n)
        {
            if (p[j]<p[i])
            {
                temp = p[i];
                p[i] = p[j];
                p[j] = p[i];
            }
            j++;
        }
        
        sum += p[i];
        i++;
    }

    printf("%d\n", sum);
    


    
    

    return 0;
}