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 Kotlin (1.0.0) (judge verdict: Runtime Error)

import java.io.BufferedReader
import java.io.IOException
import java.io.InputStreamReader
import java.io.PrintWriter
import java.util.*

/**
 * @author: pashka
 */


private fun solve() {
    var n = nextInt()
    var M = nextLong()
    var a = IntArray(n)
    for (i in 0 until n) {
        a[i] = nextInt()
    }
    a.sort()
    
    var l = 0
    var r = 3e5.toInt()
    while (r > l + 1) {
        var j = n
        var m = (l + r) / 2
        var cnt = 0L
        for (i in 0 until n) {
            // j = min : a[i] + a[j] >= m
            while (j >= 0 && a[i] + a[j - 1] >= m) {
                j--
            }
            cnt += n - j
        }
        if (cnt >= M) {
            l = m
        } else {
            r = m
        }
    }

    var j = n
    var cnt = 0L
    var sum = 0L
    var sumR = 0L
    for (i in 0 until n) {
        while (j >= 0 && a[i] + a[j - 1] >= l) {
            j--
            sumR += a[j]
        }
        cnt += n - j
        sum += (n - j) * a[i].toLong() + sumR 
    }
    sum -= (cnt - M) * l    

    println(sum)
}

fun main(args: Array<String>) {
    solve()
    out.close()
}

private var br = BufferedReader(InputStreamReader(System.`in`))
private var st: StringTokenizer? = null
private var out = PrintWriter(System.out)

@Throws(IOException::class)
private fun next(): String {
    while (st == null || !st!!.hasMoreTokens()) {
        st = StringTokenizer(br.readLine())
    }
    return st!!.nextToken()
}

@Throws(IOException::class)
private fun nextInt(): Int {
    return next().toInt()
}

@Throws(IOException::class)
private fun nextLong(): Long {
    return next().toLong()
}