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 PyPy3 (2.4.0) (judge verdict: Runtime Error)

import sys
from bisect import bisect, bisect_left
import heapq

def solve():
    input = sys.stdin.readline
    N, M = map(int, input().split())
    A = [int(a) for a in input().split()]
    A.sort()
    S = [0] * (N + 1)
    for i, a in enumerate(A): S[i + 1] = a + S[i]
 
    l, h = 1, A[N-1] * 2 + 1
    count_l = 0
    while h - l > 1: #最小値がｍ以上の要素をM個用意できるか
        m = (h + l) // 2
        count = 0
        for i, a in enumerate(A): count += N - bisect_left(A, m - a)
        if count >= M: 
            l = m
            count_l = count
        else: h = m

    total = 0
    count = 0
    Q = []
    heapq.heapify(Q)
    min_out = 10000000
    for i, a in enumerate(A):
        j = bisect_left(A, l - a)  
        total += a * (N - j) + (S[N] - S[j])
        count += N - j
        min_out = min(min_out, a + A[j])

    if count > M:
        total -= min_out * (count - M)

    print(total)

    return 0

if __name__ == "__main__":
    solve()