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 <stdio.h>
using namespace std;
#define ll long long
const int maxi=2000000000;
const int maxq=1000000000;
const ll mod=1000000007LL;
#define f(i,a,b) for(int  i=(a); i<(b);++i)
/*
unsigned int countSetBits(unsigned int n)
{
    unsigned int count = 0;
    while (n) {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
*/

/*
const int maxx=9000000;
void SieveOfEratosthenes(vector<int> &primes){
    bool IsPrime[maxx+1];
    memset(IsPrime, true, sizeof(IsPrime));
    for(int p=2;p*p<=maxx ;p++){
        if(IsPrime[p]){
            for(int i=p*p;i<maxx ;i+=p) IsPrime[i]=false;
        }
    }
    f(p,2,maxx +1){
        if(IsPrime[p]) primes.push_back(p);
    }
    }
*/
//int b[100];
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n,k,sum=0;
    cin>>n>>k;
    int a[n];
    f(i,0,n) cin>>a[i];
    sort(a,a+n);
    f(i,0,k) sum+=a[i];
    cout<<sum<<endl;
    return 0;
}