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)

from bisect import bisect, bisect_left
from itertools import accumulate


def check(x):
    count = 0
    for a in aaa:
        if x <= a:
            count += n
        else:
            count += n - bisect_left(aaa, x - a)
    # print(x, count)
    return count >= m


n, m = map(int, input().split())
aaa = list(map(int, input().split()))
aaa.sort()
acc = [0] + list(accumulate(aaa))
sum_a = acc[-1]

l, r = 0, 200001
while l + 1 < r:
    mid = (l + r) // 2
    if check(mid):
        l = mid
    else:
        r = mid
# print(l, r)

# 合計がlより大きい組み合わせの和と組数を求める→Mに足りない分だけ合計がlの組を採用
ans = 0
count = 0
for a in aaa:
    if l <= a:
        ans += sum_a + a * n
        count += n
    else:
        omit = bisect(aaa, l - a)
        if omit < n:
            ans += sum_a - acc[omit] + a * (n - omit)
            count += n - omit
        # print(a, omit, ans, count)
ans += l * (m - count)
print(ans)