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 Lua (5.3.2) (judge verdict: Accepted)

local n, m = io.read("*n", "*n")
local t = {}
for i = 1, n do
  t[i] = io.read("*n")
end
table.sort(t)
local tsum = {t[1]}
for i = 2, n do
  tsum[i] = tsum[i - 1] + t[i]
end

local function solve(x)
  local cnt = 0
  local pos = 1
  for i = n, 1, -1 do
    while pos <= n and t[pos] + t[i] < x do
      pos = pos + 1
    end
    if n < pos then break end
    cnt = cnt + n + 1 - pos
  end
  return m <= cnt
end

local min, max = 0, 200001
while 1 < max - min do
  local mid = (min + max) // 2
  if solve(mid) then
    min = mid
  else
    max = mid
  end
end
do
  local score = 0
  local cnt = 0
  local pos = 1
  for i = n, 1, -1 do
    while pos <= n and t[pos] + t[i] < min do
      pos = pos + 1
    end
    if n < pos then break end
    cnt = cnt + n + 1 - pos
    score = score + (n + 1 - pos) * t[i] + tsum[n]
    if 1 < pos then
      score = score - tsum[pos - 1]
    end
  end
  score = score - min * (cnt - m)
  print(score)
end