Problem:
Score : 500 points

Problem Statement

Takahashi has come to a party as a special guest.
There are N ordinary guests at the party. The i-th ordinary guest has a power of A_i.

Takahashi has decided to perform M handshakes to increase the happiness of the party (let the current happiness be 0).
A handshake will be performed as follows:

Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same).

Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y.

However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold:

Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q).

What is the maximum possible happiness after M handshakes?

Constraints

1 \leq N \leq 10^5

1 \leq M \leq N^2

1 \leq A_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 A_2 ... A_N

Output

Print the maximum possible happiness after M handshakes.

Sample Input 1

5 3
10 14 19 34 33

Sample Output 1

202

Let us say that Takahashi performs the following handshakes:

In the first handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 4.

In the second handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 5.

In the third handshake, Takahashi shakes the left hand of Guest 5 and the right hand of Guest 4.

Then, we will have the happiness of (34+34)+(34+33)+(33+34)=202.

We cannot achieve the happiness of 203 or greater, so the answer is 202.

Sample Input 2

9 14
1 3 5 110 24 21 34 5 3

Sample Output 2

1837

Sample Input 3

9 73
67597 52981 5828 66249 75177 64141 40773 79105 16076

Sample Output 3

8128170

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef pair<int, int> P;
#define REP(i, n) for (int i=0; i<(int)(n); ++i)

template<class T> inline bool chmin(T &a, T b) {
  if(a>b) {
    a=b;
    return true;
  }
  return false;
}
template<class T> inline bool chmax(T &a, T b) {
  if(a<b){
    a=b;
    return true;
  }
  return false;
}
const ll LINF = 1LL << 60;
int dx[] = {0,1,0,-1};
int dy[] = {-1,0,1,0};


int main() {
  int N, M; cin >> N >> M;
  int a[100010];
  REP(i, N) {
    cin >> a[i];
  }
  sort(a, a+N, greater<int>());
  int ans[200010];
  int size=0;
  for(int i=0; i<N; i++) {
    for(int j=0; j<N; j++) {
      ans[size] = a[i] + a[j];
      size++;
    }
  }
  sort(ans, ans+size, greater<int>());
  int answer = 0;
  for(int i=0; i<M; i++){
    answer += ans[i];
  }
  cout << answer <<endl;

}