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 Python (3.4.3) (judge verdict: Accepted)

#!/usr/bin/env python3
import sys
from bisect import bisect_left
from itertools import accumulate

INF = float("inf")


def solve(N: int, M: int, A: "List[int]"):
    A.sort()

    # 幸福度X以上を得られる握手をカウントし、M以上となる境界を探す
    # Xが小さい方が握手の個数は多く、Xが大きい方が握手の個数は少ない
    left = 0
    right = 3*(10**5)
    while right - left > 1:
        X = (right + left)//2

        # 握手の個数は、左手を固定した上で、右手をにぶたんする。
        tot = 0
        for left_hand in A:
            tot += N-bisect_left(A, X-left_hand)

        if tot < M:
            right = X
        else:
            left = X

    # print(left, right)
    # 幸福度rightを得られる握手はM個未満
    # 幸福度leftを得られる握手はM個以上

    # 幸福度rightを得られる握手をすべてカウントした上で、
    # 不足する個数だけ、幸福度leftを加える
    happy = 0

    acc = [0]+list(accumulate(A))
    tot = 0
    for left_hand in A:
        i = bisect_left(A, right-left_hand)
        tot += N-i
        happy += (acc[N]-acc[i]) + left_hand*(N-i)

    happy += (M-tot)*left
    print(happy)

    return


def main():

    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    M = int(next(tokens))  # type: int
    A = [int(next(tokens)) for _ in range(N)]  # type: "List[int]"
    solve(N, M, A)


if __name__ == '__main__':
    main()