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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lvector vector<ll>
#define lque queue<ll>
#define lpque priority_queue<ll>
#define dlpque priority_queue<ll,lvector,greater<ll>>
#define P pair<ll,ll>
#define ALL(a) (a).begin(),(a).end()
#define rep(i,n) for(ll (i)=0; (i)<(n); ++(i))
#define print(a) cout << (a) << endl

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n,m,ans=0;cin>>n>>m;
  lvector A(n,0),S(n,0);
  rep(i,n) cin>>A[i];sort(ALL(A),greater<ll>());
  ll xmin=0,xmax=2e5,x,shakes,k;
  while(xmax-xmin>1) {
    shakes=0;k=n-1;x=(xmax+xmin)/2;
    for(ll i=0;i<n;++i) {
      for(ll j=k;j>-1;--j) {
        if(A[j]>x-A[i]) break;
        k--;
      }
      shakes+=k+1;
    }
    if(shakes<m) xmax=x;
    else xmin=x;
  }
  S[0]=A[0];rep(i,n-1) S[i+1]+=S[i]+A[i+1];
  shakes=0;k=n-1;
  rep(i,n) {
    for(ll j=k;j>-1;--j) {
      if(A[j]>x-A[i]) break;
      k--;
    }
    if(k!=-1) ans+=S[k]+(k+1)*A[i];
    shakes+=k+1;
  }
  ans+=(m-shakes)*x;
  print(ans);
  return 0;
}