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: Time Limit Exceeded)

#include <stdio.h>
  
int main(void)
{
 
  int i, j, tmp;
  
 
  /* 数値を格納する配列 */
  int number[1000];
 
  /* 数値の総数を入力 */
  int N;
  scanf("%d", &N);
   /* Kを入力 */
  int s, K, total;
  scanf("%d",&K);
 
  /* 配列に格納する数値を入力 */
  printf("%d個の数値を入力 \n", N);
  for (i=0; i<N; ++i)
    scanf("%d", &number[i]);
 
  /* 数値を昇順にソート */
  for (i=0; i<N; ++i) {
    for (j=i+1; j<N; ++j) {
      if (K > number[i] > number[j]) {
        tmp =  number[i];
        number[i] = number[j];
        number[j] = tmp;
      }
    }
  }
 
  /* 計算 */
  for (s=0; s<K; ++i) {
    total = number[s]+number[s+1];
  }
    
  /* 昇順ソートした数値を出力 */
  printf("%d", total);
  
}