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>
using namespace std;
#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define For(i,n) for(ll i=0; i<n; ++i)
#define all(v) (v).begin(), (v).end()
#define ff first
#define ss second
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define deb(x) cout<<"For debug : "<<x<<endl;
typedef long long ll;
#define ld long double

void solve()
{
    int n,k;
    cin>>n>>k;
    vector<int> v(n);
    For(i,n) cin>>v[i];

    sort(v.begin(),v.end());
    int sum=0;
    For(i,k) sum+=v[i];
    cout<<sum<<endl;
}

int main(){
    optimize();
    ll T=1;
    //cin>>T;
    while(T--)
    {
        solve();
    }
    return 0;
}