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++ (Clang 10.0.0) (judge verdict: Accepted)

//#include <bits/stdc++.h>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define fast_io()                \
    ios::sync_with_stdio(false); \
    std::cin.tie(0);
#define mod 998244353
typedef long long ll;
typedef long double ld;
using namespace std;

int main()
{
    fast_io();
    int k, n;
    int p[1001];
    cin >> n >> k;
    for (int i = 0; i < n; i++)
        cin >> p[i];
    sort(p, p + n);
    ll ans = 0;
    for (int i = 0; i < k; i++)
        ans += p[i];
    cout << ans;
    return 0;
}