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<bits/stdc++.h>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<int,int> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const LL mod=1000000007;
const LL LINF=1LL<<60;
const int INF=1<<30;
int dx[]={1,0,1,-1,1,-1,-1,0};
int dy[]={0,1,1,-1,-1,1,0,-1};


int main(){
    LL n,m;cin >> n >> m;
    vector<LL> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    sort(ALL(a));
    LL l = 0,r = LINF;
    while(r - l > 1){
        LL mm = (r - l) / 2 + l;
        LL sum = 0;
        for (int i = 0; i < n; i++) {
            LL res = lower_bound(ALL(a),mm - a[i]) - a.begin();
            LL r = (LL)a.size() - res;
            sum += r;
        }
        if(m <= sum) l = mm;
        else r = mm;
    }
    vector<LL> cum(n+1,0);
    sort(RALL(a));
    for (int i = 0; i < n; i++) {
        cum[i+1] = cum[i] + a[i];
    }
    sort(ALL(a));
    LL ans = 0;
    LL ss = 0;
    for (int i = 0; i < n; i++) {
        LL res = lower_bound(ALL(a),l - a[i]) - a.begin();
        LL r = (LL)a.size() - res;
        ans += r * a[i] + cum[r];
        ss += r;
    }
    cout << ans - (ss - m) * l << endl;
    return 0;
}