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: Runtime Error)

from itertools import accumulate as ac
def isok(arg):
    x = 0
    for i in a:
        if arg - i >= 0:
            x += cs[arg-i]
        else:
            x += n
    if x >= m:
        return True
    else:
        return False

def bisect(ng, ok):
    while (abs(ok - ng) > 1):
        mid = (ok + ng) // 2
        if isok(mid):
            ok = mid
        else:
            ng = mid
    return ok

n,m = map(int,input().split())
a = list(map(int,input().split()))
cs = [0]*(10**5+1)
for i in a:
    cs[i] += 1
for i in range(10**5):
    cs[10**5-i-1] += cs[10**5-i]
h = bisect(10**5*2,0)
count = 0
ans = 0
for i in a:
    num = max(0,h-i+1)
    ans += 2*i*cs[num]
    count += cs[num]
ans += (m-count)*h
print(ans)