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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>

void swap(int *x,int *y){
    int z;
    z=*x;
    *x=*y;
    *y=z;
}

int main(void){
    int N,K;
    int p[1024];
    int i,m,n,l;
    int sum=0;
    scanf("%d",&N);
    scanf("%d",&K);
    for(i=0;i<N;i++){
        scanf("%d",&p[i]);
    }
    
    //バブルソート
    for(m=0;m<N;m++){
        for(n=m+1;n<=N;n++){
            if(p[m]>=p[n]){
                swap(&p[m],&p[n]);
            }
        }
    }

    printf("並び替えると");
    for(i=0;i<N;i++){
        printf("%d ",p[i]);
    }
    printf("\n");

    for(l=0;l<=K;l++){
        sum+=p[l];
    }
    printf("%d",sum);
    return 0;
}