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 <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdint>
#include <cstdio>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;
// }}}

using ll = long long;

ll N, M;
ll A[101010];
ll B[101010];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> N >> M;
    for (int i = 0; i < N; ++i) cin >> A[i];
    sort(A, A+N);
    reverse(A, A+N);

    for (int i = 0; i < N; ++i) A[i] = -A[i];
    B[0] = 0;
    for (int i = 1; i <= N; ++i) B[i] = (-A[i-1]) + B[i-1];

    ll lo = 0, hi = 1000000000000LL; 
    while (lo+1<hi) {
        ll u = (lo + hi) / 2;
        ll count = 0;
        for (int i = 0; i < N; ++i) {
            int idx = upper_bound(A, A+N, - (u + A[i])) - A;
            count += idx;
        }
        if (count >= M) lo = u;
        else hi = u;
    }

    ll ans = 0;
    ll count = 0;
    for (int i = 0; i < N; ++i) {
        int idx = upper_bound(A, A+N, - (lo + A[i])) - A;
        ans += idx * (-A[i]) + B[idx];
        count += idx;
    }
    assert(count >= M);
    cout << ans - (count - M) * lo << endl;

    return 0;
}