Problem:
Score : 200 points

Problem Statement

Snuke has N sticks.
The length of the i-th stick is l_i.

Snuke is making a snake toy by joining K of the sticks together.

The length of the toy is represented by the sum of the individual sticks that compose it.
Find the maximum possible length of the toy.

Constraints

1 \leq K \leq N \leq 50

1 \leq l_i \leq 50

l_i is an integer.

Input

Input is given from Standard Input in the following format:

N K
l_1 l_2 l_3 ... l_{N}

Output

Print the answer.

Sample Input 1

5 3
1 2 3 4 5

Sample Output 1

12

You can make a toy of length 12 by joining the sticks of lengths 3, 4 and 5, which is the maximum possible length.

Sample Input 2

15 14
50 26 27 21 41 7 42 35 7 5 5 36 39 1 45

Sample Output 2

386

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for(ll i=0;i<(n);i++)
#define rrep(i,n) for(ll i = 1; i <= (n); ++i)
#define drep(i,n) for(ll i = (n)-1; i >= 0; --i)
#define all(v) v.begin(),v.end()
#define len(x) (ll)(x).length()
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define pb push_back
#define sz(x) (ll)(x).size()
#define v(T) vector<T>
#define vv(T) vector<vector<T>>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
ll gcd(ll a,ll b){if(a%b==0){return b;}else{return(gcd(b,a%b));}}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
const int INF=1e9;
const ll MX = 1e18;
const ll MOD=INF+7;
const int di[] = {-1,0,1,0};
const int dj[] = {0,-1,0,1};
const double PI=acos(-1);
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
#define YN {puts("YES");}else{puts("NO");}
const ll MAX = 1000001;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {fac[0] = fac[1] = 1;finv[0] = finv[1] = 1;inv[1] = 1;for (ll i = 2; i < MAX; i++){fac[i] = fac[i - 1] * i % MOD;inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;finv[i] = finv[i - 1] * inv[i] % MOD;}}
long long COM(ll n, ll k){if (n < k) return 0;if (n < 0 || k < 0) return 0;return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;}
ll llpow(ll n,ll i){ll cnt=n;  for(ll j=0;j<i-1;j++){n*=cnt;}return n;}

int main() {
  int n,k;
  cin>>n>>k;
  vi l(n);
  rep(i,n) cin>>l[i];
  sort(all(l));
  reverse(all(l));
  int ans=0;
  rep(i,k){
    ans+=l[i];
  }
  cout<<ans<<endl;
}