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;

#define debug(x) cout <<"DEBUG"<<" "<< #x << ":" << x << '\n'

//↓0-originか1-originでn回繰り返し
#define rep(i,n) for(int i = 0; i < ((int)(n)); i++)   //0-origin昇順
#define rep1(i,n) for(int i = 1; i <= ((int)(n)); i++) //1-origin昇順
#define rrep(i,n) for(int i = ((int)(n)-1); i >= 0; i--)  //0-origin降順
#define rrep1(i,n) for(int i = ((int)(n)); i >= 1; i--)   //1-origin降順
//rep2 -> 第二引数 m から n 回繰り返し ex) m=5 n=3 なら i=5,i=6,i=7 まで
#define rep2(i,m,n) for(int i = ((int)(m)); i < ((int)(n))+((int)(m)); i++)
#define SIZE(x) ((int)((x).size()))
#define all(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vii; //2次元配列
typedef vector<ll> vll;
typedef vector<string> vs;
typedef pair<int, int> PII;
template<class T,class U> inline bool chmax(T &a, const U &b){if(a<b){a=b;return 1;}return 0;}
template<class T,class U> inline bool chmin(T &a, const U &b){if(a>b){a=b;return 1;}return 0;}

/* 円周率は M_PI を使おうね */
const ll infll = (1LL<<62)-1;
const int inf = (1<<30)-1;
int mod = 1000000007;

/*FUNCs=================================================*/
vi move1 = {0,0,1,-1};
vi move2 = {1,-1,0,0};

/*MAIN==================================================*/
signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);   //cin cout 高速化
    //cout << fixed << setprecision(10);

    ll n, m; cin >> n >> m;
    vll a(n); rep(i,n) cin >> a[i];
    sort(all(a));
    reverse(all(a));
    priority_queue<ll, vll, less<ll>> aa;
    for(int i = 0; i < n; i++) {
        for(int j = i; j < n; j++) {
            aa.emplace(a[i]+a[j]);
            if(i!=j) aa.emplace(a[i]+a[j]);
        }
    }
    ll ans = 0;
    ll cnt = 0;
    while(aa.size()) {
        if(cnt >= m) break;
        cnt++;
        ans += aa.top(); aa.pop();
    }
    cout << ans << '\n';
}