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: Accepted)

#include<stdio.h>
//小さい順にソート
int c[1002];
void merge(int i,int j,int *s){
  if(i!=j){
    merge(i,(i+j)/2,s);
    merge((i+j)/2+1,j,s);
    int a=i,b=(i+j)/2+1;
    int d;
    for(d=i;d<=j;d++){
      if((i+j)/2<a){
        c[d]=s[b];
        b++;
      }
      else if(j<b){
        c[d]=s[a];
        a++;
      }
      else if(s[a]<s[b]){
        c[d]=s[a];
        a++;
      }
      else{
        c[d]=s[b];
        b++;
      }
    }
    for(d=i;d<=j;d++)
      s[d]=c[d];
  }
}

int main(){
  int n,k,i,count=0;
  scanf("%d%d",&n,&k);
  int p[n+1];
  for(i=0;i<n;i++)
    scanf("%d",&p[i]);
  merge(0,n-1,p);
  for(i=0;i<k;i++)
    count+=p[i];
  printf("%d\n",count);
  return 0;
}