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>
#define ull unsigned long long
#define int long long
#define ll long long
#define fr(i, n) for (int i = 0; i < n; i++)
#define frf(i, j, n) for (int i = j; i <= n; i++)
#define frd(i, n) for (int i = n; i >= 0; i--)
#define mp(i, j) make_pair(i, j)
#define pb(x) push_back(x)
using namespace std;
// typedef vector<vector<ll> > matrix;
// const ll S = 3e5 + 20;
// const ll N = 1e6 + 1;
// const ll PRIME = 1e9 + 7;
// const ll MAX = 1e6;

void solve()
{
    int n, k;
    cin >> n >> k;
    int a[n];
    fr(i, n)
    {
        cin >> a[i];
    }
    sort(a, a + n);
    int ans = 0;
    fr(i, k)
    {
        ans += a[i];
    }
    cout << ans;
}
int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    int t;
    // cin >> t;
    t = 1;
    while (t--)
    {
        solve();
    }
}