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;
using ll = long long;

ll solve() {
    ll N, M;
    cin >> N >> M;
    vector<ll> A(N);
    for ( int i = 0; i < N; i++ ) {
        cin >> A[i];
    }
    sort(A.begin(), A.end(),greater<>());

    // Ax+Ayがaa以上となる組み合わせcm
    auto get_m = [&](ll aa) {
        ll cm = 0;
        for ( ll i = 0; i < N; i++ ) {
            if ( A[i] + A[0] < aa ) break; 
            ll p = upper_bound(A.begin(), A.end(), aa - A[i], greater<>()) - A.begin();
            //cm += max(p-i+1, 0LL);
            cm += p;
        }
        return cm;
    };
    // Ax+Ayがaa以上となる組み合わせcmがM未満 r:ok
    auto chk_m = [&](ll aa) {
        ll cm = get_m(aa);
        return cm < M;
    };

    ll l = 0, r = 2*A[0]+1;
    while ( r - l > 1 ) {
        ll m = ( r + l ) / 2;
        if ( chk_m(m) ) r = m;
        else l = m;
    }
    // r:A+A最低値+1
    ll cu = get_m(r);

    vector<ll> ca(N+1);
    for ( int i = 0; i < N; i++ ) {
        ca[i+1] = ca[i] + A[i];
    }
    ll sum_a = 0;
    for ( ll i = 0; i < N; i++ ) {
        if ( A[i] + A[0] < r ) break;
        ll p = upper_bound(A.begin(), A.end(), r - A[i], greater<>()) - A.begin();
        // sum_a += ca[p+1] - ca[i];
        // sum_a += A[i] * (p-i+1);
        sum_a += ca[p];
        sum_a += A[i] * (p);
    }

    sum_a += l * (M - cu);
    return sum_a;
}

int main() {
    auto ans = solve();
    cout << ans << "\n";
    return 0;
}