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: Wrong Answer)

import sys
import math
from collections import defaultdict

sys.setrecursionlimit(10**7)
def input():
    return sys.stdin.readline()[:-1]

mod = 10**9 + 7

def I(): return int(input())
def II(): return map(int, input().split())
def III(): return list(map(int, input().split()))
def Line(N,num):
    if N<=0:
        return [[] for _ in range(num)]
    elif num==1:
        return [I() for _ in range(N)]
    else:
        read_all = [tuple(II()) for _ in range(N)]
        return map(list, zip(*read_all))

#################

import cmath
pi = cmath.pi
exp = cmath.exp

def convolution(a,b):
    
    def fft(a,sz,inv=False):
        tmp = [0]*sz
        mask = sz-1
        p = 0
        i = sz>>1
        sign = 1 if inv else -1
        e = exp(2j*pi*sign/sz)

        while i:
            if p&1:
                cur,nex = tmp,a
            else:
                cur,nex = a,tmp
            ei = e**i
            w = 1
            for j in range(0,sz,i):
                for k in range(i):
                    nex[j+k] = cur[((j<<1)&mask)+k] + w*cur[(((j<<1)+i)&mask)+k]
                w *= ei
            p += 1
            i >>= 1

        if p&1:
            a,tmp = tmp,a
        if inv:
            a = list(map(lambda x: x/sz, a))
        return a
    
    sz = 1<<(len(a)+len(b)-2).bit_length()
    a = a + [0]*(sz-len(a))
    b = b + [0]*(sz-len(b))

    fa = fft(a,sz)
    fb = fft(b,sz)
    fafb = [fai*fbi for fai,fbi in zip(fa,fb)]
    ab = fft(fafb,sz,inv=True)
    return [round(x.real) for x in ab[:len(a)+len(b)-1]]

N,M = II()
A = III()

h = [0]*(max(A))
for a in A:
    h[a-1] += 1

conv = [0,0] + convolution(h,h)
ans = 0
for k in range(2,2*max(A)+1)[::-1]:
    if conv[k]<M:
        ans += k*conv[k]
        M -= conv[k]
    else:
        ans += k*M
        break
        
print(ans)