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 Ruby (2.3.3) (judge verdict: Accepted)

MOD=10**9+7
cnt=0; sum=0; prev=nil; can=true; h=Hash.new(0)
def gs() gets.chomp end
def gi() gets.to_i end
def gsmi() gets.chomp.split.map(&:to_i) end

def desc(ar) ar.sort!{|x,y|y<=>x} end
def min(a,b) a<=b ? a : b end
def max(a,b) a>=b ? a : b end
def sum(ar) ar.inject(:+) end

def C(a,b) b==0||a==b ? 1 : (b=a-b if a/2<b;(a-b+1..a).inject(:*)/(1..b).inject(:*)) end

def puts_yesno(can) puts(can ? 'Yes' : 'No') end
def putsend(s) puts s; exit end
def debug(k,v) puts "#{k}:#{v}" end
########### ( ˘ω˘ )ｽﾔｧ… ###########
n,m=gsmi
a=gsmi.sort

# "増える量がX以上となるのがmとなる" Xを二分探索で探す O(N(logN)^2)
# lは条件を満たす、rは条件を満たさない(X=lとなる)
l=0
r=2*10**5+1
while l+1!=r do
  cen=(l+r)/2
  #puts "l:#{l},r:#{r},cen:#{cen}"
  # TODO必要ならbreak文
  cnt=0
  n.times do |i|
    i2=cen-a[i]
    idx=a.bsearch_index{|x|x>=i2}
    next if idx.nil?

    c=(n-1)-idx+1
    cnt+=c
    # M以上組み合わせがあるならbreak
    break if cnt>=m
  end

  if cnt>=m
    #OKなので右へ
    l=cen
  else
    r=cen
  end
end

# 累積和
ruisum=[0]
n.times do |i|
  ruisum<< ruisum[-1]+a[i]
end
#puts ruisum.to_s

# 合計を求める
sum=0
cnt=0
n.times do |i|
  i2=l-a[i]
  idx=a.bsearch_index{|x|x>=i2}
  next if idx.nil?

  c=(n-1)-idx+1
  cnt+=c

  rui=ruisum[n]-ruisum[idx]
  s=c*a[i]+rui
  sum+=s
end

# 余分に加えた分を補正
puts sum-l*(cnt-m)