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: Time Limit Exceeded)

import cmath
pi = cmath.pi

def expD(t):
  nexp, pexp = {}, {}
  while t != 0:
    pexp[t] = cmath.exp(2j*pi/t)
    nexp[t] = cmath.exp(-2j*pi/t)
    t //= 2
  return nexp, pexp
def powerlimit(x):
  a = 1
  while a < x:
    a *= 2
  return a

N, M = map(int, input().split())
A = list(map(int, input().split()))
m = max(A)
X = [0 for i in range(m+1)]
for i in A:
  X[i] += 1
limit = powerlimit(2*(m+1))
X = X[:] + [0 for _ in range(limit-m-1)]

nexp, pexp = expD(limit)

def fft(f, n):
  hn = n//2
  if n == 1:
    return f
  elif n == 2:
    return [f[0] + f[1], f[0] - f[1]]
  else:
    f0 = [f[2*_] for _ in range(hn)]
    f1 = [f[2*_+1] for _ in range(hn)]
    f0 = fft(f0, hn)
    f1 = fft(f1, hn)
    grow = 1
    fund = nexp[n]
    for i in range(hn):
      right = grow * f1[i]
      f[i] = f0[i] + right
      f[i+hn] = f0[i] - right
      grow *= fund
    return f
def ifft(f, n):
  hn = n//2
  if n == 1:
    return f
  elif n == 2:
    return [f[0] + f[1], f[0] - f[1]]
  else:
    f0 = [f[2*_] for _ in range(hn)]
    f1 = [f[2*_+1] for _ in range(hn)]
    f0 = ifft(f0, hn)
    f1 = ifft(f1, hn)
    grow = 1
    fund = pexp[n]
    for i in range(hn):
      right = grow * f1[i]
      f[i] = f0[i] + right
      f[i+hn] = f0[i] - right
      grow *= fund
    return f
  
F = fft(X, limit)
V = [F[i]*F[i] for i in range(limit)]
E = ifft(V, limit)
for i in range(limit):
  E[i] = int(0.5+E[i].real/limit)
cnt = 0
S = 0
point = limit - 1

while cnt < M:
  Ep = E[point]
  if Ep != 0:
    if cnt + Ep > M:
      S += (M - cnt) * point
      cnt = M
    else:
      S += point * Ep
      cnt += Ep
  point -= 1
print(S)