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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
//#include "atcoder/all"
typedef long long int ll;
using namespace std;
// using namespace atcoder;
#define MAXN 100001
vector<ll> vec;
ll sum[MAXN];
int main() {
    ll n, m;
    cin >> n >> m;
    sum[0] = 0;
    for (int i = 1; i <= n; i++) {
        ll a;
        cin >> a;
        vec.push_back(a);
    }
    sort(vec.begin(), vec.end());
    for (int i = 1; i <= n; i++) {
        sum[i] = sum[i - 1] + vec[n - i];
    }
    ll Ma = 200001;
    ll Mi = 1;
    // k以上の幸福が m パターンある
    while (Ma - Mi > 1) {
        ll med = (Ma + Mi) / 2;
        ll count = 0;
        // cout << "med " << med << endl;
        for (int i = 0; i < n; i++) {
            auto it = lower_bound(vec.begin(), vec.end(), med - vec[i]);
            int ok;
            if (it == vec.end()) {
                ok = 0;
            }
            else {
                ok = n - (it - vec.begin());
            }
            // cout << "i " << i << ok  << endl;
            count += ok;
        }
        // cout << count << endl;
        if (count < m) {
            Ma = med;
        }
        else {
            Mi = med;
        }
    }
    // cout << Mi << endl;
    ll ret = 0;
    ll co = 0;
    for (int i = 0; i < n; i++) {
        auto it = lower_bound(vec.begin(), vec.end(), Mi - vec[i] + 1);
        int ok;
        if (it == vec.end()) {
            ok = 0;
        }
        else {
            ok = n - (it - vec.begin());
        }
        co += ok;
        ret += sum[ok] + vec[i] * ok;
    }
    ret += Mi * (m - co);
    cout << ret << endl;
    return 0;
}