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: 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))

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

# FFT
# use python3
# a = np.array([a1,a2,a3]), b= np.array([b1,b2,b3])
# c = np.array([a1b1,a1b2+a2b1,a1a3+a2b2+a3b1,a2b3+a3b2,a3b3])
import numpy as np
def Convolution(a,b):
    bit = (len(a)+len(b)-2).bit_length()
    L = 2**bit
    fa,fb = np.fft.rfft(a,L), np.fft.rfft(b,L)
    c = np.rint(np.fft.irfft(fa*fb,L)).astype(np.int64)
    return c[:len(a)+len(b)-1]

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

print('time test')

h = np.zeros(max(A))
for a in A:
    h[a-1] += 1

conv = np.append([0,0],Convolution(h,h))
ans = 0
count = 0
'''
for k in range(2,2*max(A)+1)[::-1]:
    if conv[k]:
        num = min(M-count,conv[k])
        count += num
        ans += k*num
    if count==M:
        break
'''
print(ans)