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: Wrong Answer)

// ヒント: atcoderの億マス計算と同じ?らしい
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#define rep(i, start, end) for (int i = (int)start; i < (int)end; ++i)
#define rrep(i, start, end) for (int i = (int)start - 1; i >= end; --i)
#define iter(i, c) for (auto i = (c).begin(); i != (c).end(); ++i)
#define riter(i, c) for (auto i = (c).rbegin(); i != (c).rend(); ++i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
using namespace std;
using ll = long long;
using P = pair<ll, ll>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N, M;
    cin >> N >> M;
    vector<ll> A(N);
    rep(i, 0, N) cin >> A[i];
    sort(all(A));
    rep(i, 0, N) cout << A[i] << ' ';
    cout << endl;
    vector<ll> cumsum(N + 1, 0);
    rep(i, 0, N) {
        cumsum[i + 1] = A[i] + cumsum[i];
    }
    ll l = 2, r = 202020;
    while (r - l > 1) {
        ll mid = (l + r) / 2;
        ll cnt = 0;
        rep(i, 0, N) {
            auto itr = lower_bound(all(A), mid - A[i]);
            cnt += (ll)(A.end() - itr);
        }
        if (cnt >= M) {
            l = mid;
        }
        else {
            r = mid;
        }
    }
    ll cnt = 0;
    ll ans = 0;
    rep(i, 0, N) {
        auto itr = lower_bound(all(A), l - A[i]);
        ll now_cnt = (ll)(A.end() - itr);
        cnt += now_cnt;
        ans += cumsum[N] - cumsum[(int)(itr - A.begin())];
        ans += A[i] * now_cnt;
    }
    ans -= l * (cnt - M);
    cout << ans << endl;
    return 0;
}