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<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define MOD  1000000007
#define ll long long
#define int long long
# define INF 1000005
#define ar array
# define fr first
# define sc second
#define wt(x) int x;cin>>x;while(x--)
#define watch(x) cerr << "\n" << (#x) << " is " << (x) << endl
//#define cerr if(false)cerr
typedef tree<pair<int,int>,null_type,less<pair<int,int>>,rb_tree_tag,tree_order_statistics_node_update> pbds;

void  c_p_c(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
 
#ifndef ONLINE_JUDGE
  freopen("input1.txt","r",stdin);
  freopen("output1.txt","w",stdout);
#endif
 }
 
int power(int a,int b,int mod){
    a=a%mod;
    if(a==0){
        return 1;
    }
    int res=1;
    while(b){
        if(b&1){
            res=(res*a)%mod;
        }
        b>>=1;
        a=(a*a)%mod;
    }
    return res;
 }
 
int  MMI(int p){
    return power(p,MOD-2,MOD);
}


int32_t main()
{
    c_p_c();
    
       int n,k;
       cin>>n>>k;
       int s=0;
       vector<int>a(n);
       for(int i=0;i<n;i++)cin>>a[i];
       nth_element(a.begin(),a.begin()+k,a.end());
       for(int i=0;i<k;i++){
        s+=a[i];
       }
       cout<<s;


 
    
      return 0;
       }