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)

#include <bits/stdc++.h>
using namespace std;
typedef long long int64;

int main() {
    int N, M;
    cin >> N >> M;
    vector<int64> A(N), sum(N + 1);
    for(int i = 0; i < N; ++i) cin >> A[i];
    sort(A.begin(), A.end());
    for(int i = 0; i < N; ++i) sum[i + 1] = sum[i] + A[i];
    // 1 3 3 5 5 21 24 34 110
    // 幸福度が x 以上になるような握手の組み合わせの数と、そのときの幸福度の合計
    auto isOK = [&](int x) {
        int64 cnt = 0, total = 0;
        for(int i = 0; i < N; ++i) {
            int n = A.end() - lower_bound(A.begin(), A.end(), x - A[i]);
            cnt += n;
            total += sum[N] - sum[N - n];
            total += A[i] * n;
        }
        return make_pair(cnt, total);
    };
    int l = 0, r = 1e9;
    while(r - l > 1) {
        int mid = (l + r) / 2;
        if(isOK(mid).first >= M) l = mid;
        else r = mid;
    }
    auto ans = isOK(l);
    cout << ans.second - (int64)l * (ans.first - M) << '\n';
    return 0;
}