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;
using ll = long long;
using Graph = vector<vector<int>>;
const ll mod=1000000007;
const int MAX_N = 1000;         // n の最大値
 
// nCk を取得
double nCk(int n, int k) {
double res=1.0;
for(int i=0; i<n; i++){
res*=0.5;}
for(int i=0; i<k; i++){
res*=(double)(n-i);
res/=(double)(k-i);
}
return res;}
 
int main() {
ll n,k;
cin>>n>>k;
ll p[n]={};
ll counter=0;
for(ll i=0; i<n; i++){
cin>>p[i];}
sort(p,p+n);
for(ll i=0; i<k; i++){
counter+=p[i];}
cout<<counter<<endl;}