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 Scala (2.11.7) (judge verdict: Wrong Answer)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val N = sc.nextInt()
    val M = sc.nextLong()
    val A = Array.fill(N)(sc.nextLong()).sorted(Ordering.Long.reverse)

    // 左手と右手で握手できる.
    // その時, 左手で握手した人のパワー + 右手で握手した人のパワーが上昇する
    // しかし, 全く同じ握手を二回以上行えない,
    // 右手と左手を入れ替えて握手することは可能, また, 両手で握手することも可能
    // パワーが大きい順に貪欲に握手していけば良さそうな気がする
    // まず、パワーが大きい順にソート.
    // 左から, 両手で握手して, 左手で握手,
    // ただ, Mが10^5なので, このままでは間に合わない
    // 握手を行う人と, 握手を行わない人を分別する必要がある
    // 二分探索でその境界を探す
    // 握手を行う人の, 握手の組み合わせの数, 「握手する人から2つを選ぶ数」がM以上になる境界
    // 「握手する人から2つを選ぶ数」は, 「握手をする人 * 握手をする人」通りある
    // そしてその仮定の素では、選んだ全ての人と1 + 2 * (L - 1)回握手することになる, 両手は一回, 他の人と左手, 右手
    // そして探したM以上になるやつで, Mになるまで小さい順から握手を減らしていけばいい
    // うーーん、違うか
    def solve(N: Int, M: Long, A: Array[Long]): Long = {
      // 組み合わせの数が M 以上になる握手人数
      def lowerBounds(M: Long): Int = {
        def loop(left: Int, right: Int): Int =
          if (right - left > 1) {
            val mid = left + (right - left) / 2
            if (mid * mid >= M) loop(left, mid)
            else loop(mid, right)
          } else right
        loop(-1, N)
      }
      val L = lowerBounds(M)
      val MAXP = (0 until L).map(i => A(i)).sum * (2 * (L - 1) + 1)// 最大数
      val AA = A.take(L).sorted // 対象にしているやつのみ選び昇順にソートする

      println(MAXP, L)
      def loop(i: Int, rest: Long, acm: Long): Long =
        if (rest > M) {
          if (M - rest >= 3) {
            loop(i + 1, rest - 3, acm - 4 * AA(i) - 2 * AA(i+1))
          } else if (M - rest >= 2) {
            loop(i + 1, rest - 2, acm - 3 * AA(i) - 1 * AA(i+1))
          } else {
            loop(i + 1, rest - 1, acm - 2 * AA(i))
          }
        } else acm
      loop(0, L, MAXP)
    }

    println(solve(N, M, A))
  }

}