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: Accepted)

from bisect import bisect_left, bisect_right
n, m = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
# print(a)

a_cs = [0] * (n + 1)
for i in range(n):
    a_cs[i + 1] = a_cs[i] + a[n - 1 - i]


def check(x):
    # M回の握手の幸福度の最小値をx以上にできるか
    # つまり、M回の任意の握手の幸福度をx以上にできるか
    cnt = 0
    for ai in a:
        idx = bisect_left(a, x - ai)
        cnt += n - idx
    return cnt >= m


lb = 1  # True
ub = 2 * 10**5 + 5  # False
while ub - lb > 1:
    mid = (ub + lb) // 2
    if check(mid):
        lb = mid
    else:
        ub = mid

# print(lb)

ans = 0
cnt = 0
for ai in a:
    idx = bisect_left(a, lb - ai)
    ans += ai * (n - idx) + a_cs[n - idx]
    cnt += n - idx
ans -= lb * (cnt - m)
print(ans)