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 <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <string>
#include <sstream>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
using namespace std;
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
int solve(){
    int n, k, v,s=0;
    cin>>n>>k;
    priority_queue<int> q;
    while(n--){
      cin>>v;
      q.push(v);
      if(q.size()>k) q.pop();
    }
    while(!q.empty()){
      s+=q.top();
      q.pop();
    }
    return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << solve() << "\n";
  return 0;
}