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)

import bisect
N,M=map(int,input().split())
A=list(map(int,input().split()))
A.sort()

#和がx以上になる組み合わせの総数を求める関数
def pattern_num_ge_x(x):
    p=0
    for a in A:
        p+=(N-bisect.bisect_right(A, x-1-a))
    return(p)

#pattern_num_ge_x(x)がM未満になる最小のxを探す(=r)
#lはM以上になる最大のx
l=-1
r=10**10+1
while r-l>1:
    m=(r+l)//2
    if pattern_num_ge_x(m)<M:
        r=m
    else:
        l=m

#和がr以上になる組み合わせの総和を求める(累積和を使用)
S=[0]+list(reversed(A[:]))
for i in range(1,N+1):
    S[i]+=S[i-1]
ans=0
for a in A:
    idx=bisect.bisect_right(A,r-a-1)
    ans+=a*(N-idx) + S[N-idx]
    M-=(N-idx)

#回数のあまり分lを足す
ans+=l*M

print(ans)