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)

def nmx(x):
    r = 0
    for a in As:
        r += d2[max(0, x-a)]
    return -r
def bsearch(target, min_i, max_i, func):
    # func(index) <= target < func(index+1) となるindexを返す
    if func(max_i) <= target:
        return max_i
    if target < func(min_i):
        return None
    index = (max_i + min_i)//2
    while True:
        if func(index) <= target:
            if target < func(index+1):
                return index
            index, min_i = (index+1 + max_i)//2, index+1
            continue
        index, max_i = (index-1 + min_i)//2, index-1
N, M = map(int, input().split())
As = sorted(list(map(int, input().split())))
mn, mx = As[0], As[-1]

j = 0
n = N
d2 = []
for i in range(mx*2):
    while True:
        if j>=N:
            break
        if i <= As[j]:
            break
        j += 1
        n -= 1
    d2.append(n)
# d2[i] : i以上のaがいくつあるか
r = bsearch(-M, 1, 10**7, nmx)
from itertools import accumulate
aas = list(accumulate([0]+As[::-1]))
rr = 0
c = 0
for a in As[::-1]:
    if c+d2[max(0, r-a)] > M:
        rr += a*(M-c) + aas[M-c]
        break
    else:
        rr += a*d2[max(0, r-a)] + aas[d2[max(0, r-a)]]
    c += d2[max(0, r-a)]
print(rr)