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: Accepted)

#include<stdio.h>

#define N 1024

//sort and  upper k
int main(){
int n,k;
int p[N]={0};
int tmp=0;
  
  scanf("%d %d",&n,&k);
  for(int i=0;i<n;i++)scanf("%d",&p[i]);
  //for(int i=n-1;i>=0;i--) printf("%d\n",p[i]);puts("");
  
  for(int i=0; i<n; i++){          //n- i  is ng make it simple 
      for(int j=i+1; j<n; j++){ //for(int j=; j<n-i; j++){//for(int j=i+1; j<n-i; j++){ //for(int j=n-1;n>=0;j--){
        if( p[i] > p[j] ){
          tmp=p[i];
          p[i]=p[j];
          p[j]=tmp;
        }  
                
      }
  }
  //for(int i=n-1;i>=0;i--) printf("%d\n",p[i]);
  //all conditon ? not || it is or
    
    for(int i=0;i<k;i++) p[1008]+=p[i];   
    printf("%d\n",p[1008]);
    
    
return 0;
}

/*
5 3
50 100 80 120 80

*/