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 __gnu_pbds;
using namespace std;
 
typedef long long int ll;
typedef unsigned long long int sll;
typedef long double ld;
#define A 1000000007
#define D 100000000000000ll
#define B 998244353ll
#define C 1000000000000000000ll
#define M 4000
#define FAST ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define oset tree<pair<ll,ll>, null_type,less_equal<pair<ll,ll>>, rb_tree_tag,tree_order_statistics_node_update> 
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define mxe(v) *max_element(v.begin(),v.end())
#define mne(v) *min_element(v.begin(),v.end())
#define bs binary_search
#define lb lower_bound
#define ub upper_bound
#define ve vector
#define br break
#define PI acos(-1)
#define subt cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n" 

int main() 
{
    FAST;
    ll n,k;
    cin>>n>>k;
    ll a[n];
    for(int i=0;i<n;i++)
        cin>>a[i];
    sort(a,a+n);
    ll ans=0;
    for(int i=0;i<k;i++)
        ans+=a[i];
    cout<<ans;
    subt;
    return 0;
}