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: Accepted)

#include <iostream>
#include <fstream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>

#define fio() cin.tie(0); ios::sync_with_stdio(false);
#define REP(i, n) for (long long i = 0; i < (n); i++) 
typedef long long ll;
static const ll MOD = 1000000007;
static const ll INF = 1000000000000000000LL;
using namespace std;

int main(){
    fio();
    ll N,M;
    cin >> N >> M;
    vector <ll> a(N);
    REP(i,N)cin >> a[i];
    sort(a.begin(),a.end());

    //組合せがM以上となるものを二分探索
    ll ok = 200001;
    ll ng = -1;
    while(ng + 1 < ok)
    {
      ll mid = (ok + ng) / 2;
      ll num = 0;
      //ここでN回すのか
      REP(i,N)
      {
        //ここ組合せになってるのか
        ll zan = mid - a[i];
        //zanが対象
        auto itr = lower_bound(a.begin(), a.end(), zan);
        num += a.end() - itr;
      }
      if(num > M)ng = mid;
      else ok = mid;
    }

    //
    ll res = 0;
    vector <ll> sum(N+1, 0);
    ll num = 0;
    //累積和
    REP(i, N)sum[i+1] = sum[i] + a[i];
    
    REP(i, N)
    {
      ll zan = ok - a[i];
      auto itr = lower_bound(a.begin(), a.end(), zan);
      num += a.end() - itr;
      ll idx = itr - a.begin();
      
      //a[i]+ zan >= okを満たすもの
      res += sum[N] - sum[idx];
      //a[i]の個数
      res += a[i] * (N - idx);
    }

    //ちょうどMにはならないので、最後に、最低スコアを足し合わせる
    if(num < M)
    {
      res += (M - num) * ng;
    }

    cout << res << endl;
    return 0;    
}