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 Pascal (FPC 2.6.2) (judge verdict: Runtime Error)


var n,m,count,i,j,j1,sum,x,tam,k: longint;
a: array[1..10000] of longint;
 procedure  qs(l,r: longint);
BEGIN
i:=l;
j:=r;
x:=a[(i+j) div 2];
repeat
while a[i]>x do inc(i);
while a[j]<x do dec(j);
if i<=j then begin tam:=a[i]; a[i]:=a[j]; a[j]:=tam; inc(i); dec(j); end;
until i>j;
if l<j then qs(l,j);
if r>i then qs(i,r);
end;
begin

read(n,k);
for i:=1 to n do read(a[i]);
qs(1,n);
for i:=1 to n do
for j:=i to n do
begin
if i=j then begin
sum:=sum+a[i]+a[j];
inc(count);   if count=k then begin write(sum);  exit; end;
end else  begin sum:=sum+(a[i]+a[j])*2; inc(count); if
count=k then begin write(sum); exit; end; end;
end;


end.