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 LuaJIT (2.0.4) (judge verdict: Accepted)

local mfl = math.floor

local function comp(a, b)
  return a < b
end

local function lower_bound(ary, x)
  local num = #ary
  if num == 0 then return 1 end
  if not comp(ary[1], x) then return 1 end
  if comp(ary[num], x) then return num + 1 end
  local min, max = 1, num
  while 1 < max - min do
    local mid = mfl((min + max) / 2)
    if comp(ary[mid], x) then
      min = mid
    else
      max = mid
    end
  end
  return max
end

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

local function solve(x)
  local cnt = 0
  for i = 1, n do
    local dst = x - a[i]
    local pos = lower_bound(a, dst)
    cnt = cnt + n + 1 - pos
  end
  return m <= cnt
end

local min, max = 0, 200001
while 1 < max - min do
  local mid = mfl((max + min) / 2)
  if solve(mid) then
    min = mid
  else
    max = mid
  end
end
local ret = 0LL
local mcnt = 0
for i = 1, n do
  local dst = min - a[i]
  local pos = lower_bound(a, dst)
  if pos == 1 then
    ret = ret + asum[n] + n * a[i]
  else
    ret = ret + asum[n] - asum[pos - 1] + (n + 1 - pos) * a[i]
  end
  mcnt = mcnt + n + 1 - pos
end
if m < mcnt then
  ret = ret - (mcnt - m) * min
end
local str = tostring(ret):gsub("LL", "")
print(str)