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 <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#include <iomanip>
#include <stack>
#include <list>
#include <map>
#include <unordered_map>
#include <chrono>
#include <numeric>
using namespace std;
using ll = long long;
const ll INF = (ll)1e18+1;
const ll DIV = 1000000007;
//#define TEST
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
#ifdef TEST
    chrono::system_clock::time_point start, end;
    start = chrono::system_clock::now();
#endif
    long N, K;
    cin >> N >> K;
    vector<long> p(N);
    for(size_t i=0;i<N; i++)
    {
        cin >> p[i];
    }
    sort(p.begin(), p.end());
    ll total = 0;
    for(size_t i=0;i<K; i++)
    {
        total += p[i];
    }
    cout << total << endl;
#ifdef TEST
    end = chrono::system_clock::now();
    cerr << static_cast<double>(chrono::duration_cast<chrono::microseconds>(end - start).count() / 1000.0) << "[ms]" << endl;
#endif
    return 0;
}