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)

/**Bismillahir Rahmanir Rahim.**/

/**
 *    author:  nasimnoob
**/

#include<bits/stdc++.h>
#include<stdlib.h>
#include<iostream>
#include<string>
#include<math.h>
using namespace std;

//----------------------------------------------------------//

typedef long long int                        ll;
typedef unsigned long long int              ull;
typedef long double                          ld;

//-----------------------------------------------------------//
#define      test                       int t; cin>>t;  while(t--)

#define      cube(x)                    (x*x*x)
#define      sqr(x)                     (x*x)

#define      sc1(x)                     scanf("%lld",&x);
#define      sc2(x,y)                   scanf("%lld %lld", &x, &y)
#define      scf1(x)                    scanf("%lf",&x)
#define      scf2(x,y)                  scanf("%lf %lf", &x, &y)

#define      srev                       reverse(s.begin(),s.end())
#define      sortstr                    sort(s.begin(), s.end());
#define      gtl(x)                     getline(cin, (x))

#define      PI                         acos(-1) // 3.1415926535897932
#define      MIN                        9999999

#define      Faster                     ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

//scanf("%[^\n]%*c", str);
//sort(s,s+n,greater<int>());

//-----------------------------------------------------------//

int main()
{
    Faster;
    //string ;
    int k , n;
    cin >> k >> n;
    int a[10000];
    for(int i = 0 ; i < k ; i++)
    {
        cin >> a[i];
    }
    sort(a,a+k);
    int sum = 0;
    for(int i = 0 ; i < n ; i++)
    {
        sum+=a[i];
    }
    cout << sum << endl;
    return 0;
}