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;
typedef long long ll;
typedef vector<int> vi;
#define all(v) (v).begin(), (v).end()
#define pb push_back
#define in(n) cin >> n
#define in2(n, m) cin >> n >> m
#define out(n) cout << n << '\n'
#define rep(i, n) for (ll i = 0; i < (n); i++)
const int inf = 1000000000;
const ll linf = (ll)inf * (ll)inf;
void solve()
{
    int n, k;
    in2(n, k);
    vi p(n);
    rep(i, n) in(p[i]);
    sort(all(p));
    int ans = 0;
    rep(i, k) ans += p[i];
    out(ans);
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    solve();
}