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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
fscanf(STDIN, "%d%d", $N, $M);
$A = array_map("intval", explode(" ", trim(fgets(STDIN))));
sort($A);

//累積和を作る
$AS[-1] = 0;
foreach($A as $i => $a){
    $AS[$i] = ($AS[$i-1] ?? 0) + $a;
}
//二分探索 -- $k以上の値の個数がMを超えるなかで最大のkを求める
//めぐる式二分探索
$ok = 0; 
$ng = 10**5*2+1;
while(abs($ok - $ng) > 1){
    $mid = intdiv($ok+$ng, 2);
    $sum = 0;
    $count = 0;
    if(isOK($mid)){
        $ok = $mid;
    }else{
        $ng = $mid;
    }
}
echo $sum - $ok * ($count - $M);
function isOK($bound){
    global $M, $A, $AS, $N, $sum, $count;
    foreach($A as $a){
        $i = lower_bound($A, $bound - $a);
        $ok = $N - $i;
        $count += $ok;
        $sum += $AS[$N - 1] - $AS[$i - 1] + $a * $ok;
    }
    return $count >= $M;
}
function lower_bound($array, $key){
    global $N;
    global $memo;
    if(!isset($memo[$key])){
        $min = -1;
        $max = $N;
        while($max - $min > 1){
            $c = intdiv($min + $max, 2);
            if($array[$c] >= $key){
                $max = $c;
            }else{
                $min = $c;
            }
        }
        $memo[$key] = $max;
    }
    return $memo[$key];
}