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 <math.h>

int main(){
    int N;
    int K;
    scanf("%d",&N);

    scanf("%d",&K);
    int p[N];
    int min[N];
    int sum=0;
    int mins;
    
    for (int i=0;i<N;i++){
        scanf("%d",&p[i]);
    }
    for (int i=0;i<K;i++){  
    min[i] = p[i];
    }
   

    for(int j=K;j<N;j++){
        for (int k=0;k<K;k++){
            if((p[j]>min[k])&&(p[j]<=min[k+1])){
                mins = min[k];
                min[k] = p[j];
                for (int x=k+1;x<K;x++){
                    min[x] = mins;
                    mins = min[x+1];
                }
            }

    
        }
    }
    
    
    for (int i=0;i<K+1;i++){  
        printf("%d",min[i]);
        sum += min[i];
    }

    printf("\n%d",sum);
    return 0;
}