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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
#define fi first
#define se second
typedef pair<ll,ll> P;
using VP = vector<P>; using VVP = vector<VP>;
using VI = vector<ll>; using VVI = vector<VI>; using VVVI = vector<VVI>;
const int inf=1e9+7;
const ll INF=1LL<<61;
const ll mod=1e9+7;

int main(){
  int i,j;
  ll n,m;
  cin>>n>>m;
  VI a(n);
  for(i=0;i<n;i++){
    cin>>a[i];
  }
  sort(a.begin(),a.end());

  ll l=-1;
  ll r=202020;
  while(r-l>1){
    ll mid=(r+l)/2;
    ll cnt=0;
    for(i=0;i<n;i++){
      cnt+=a.end()-lower_bound(a.begin(),a.end(),mid-a[i]);
    }
    if(cnt>=m){
      l=mid;
    }
    else r=mid;
  }

  ll ans=0;
  ll cnt=0;
  for(i=0;i<n;i++){
    auto it=upper_bound(a.begin(),a.end(),l-a[i]);
    while(it!=a.end()){
      ans+=*it+a[i];
      cnt++;
      ++it;
    }
  }
  ans+=l*(m-cnt);
  cout<<ans<<endl;
 
  
  
  
  return 0;
}