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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import (
	"fmt"
	"sort"
)

var cacheTotal map[int]int

func main() {
	var n, m, score int
	var power []int
	cacheTotal = make(map[int]int)
	fmt.Scanf("%d %d", &n, &m)

	for i := 0; i < n; i++ {
		var tmp int
		fmt.Scanf("%d", &tmp)
		power = append(power, tmp)
	}

	sort.Sort(sort.Reverse(sort.IntSlice(power)))
	//fmt.Printf("power: %+v\n", power)

	min, count := findByBTree(power, m)

	//fmt.Printf("min: %d count: %d\n", min, count)
	//fmt.Printf("cacheTotal: %+v\n",cacheTotal)
	score = cacheTotal[min]  //calcSum(min)
	//fmt.Printf("score: %+v\n", score)

	if m < count {
		score -= min * (count - m)
	}

	fmt.Print(score)
}

func findByBTree(array []int, target int) (int, int) {
	left := 0
	right := array[0] * 2
	mid := 0
	count := 0
	for left < right {
		mid = (left + right) / 2
		//fmt.Printf("left: %d right: %d mid:%d \n", left, right, mid)
		count = numOfElem(array, mid, target)
		//fmt.Printf("count:%d \n", count)
		if count == target {
			return mid, count
		} else if count > target {
			left = mid + 1
		} else if count < target {
			right = mid
		}
	}

	return mid, count
}

func numOfElem(array []int, min int, target int) int {
	count := 0
	sum := 0
	for _, num := range array {
		if target < count {
			break
		}
		for i := 0; i < len(array); i++ {
			tmp := num + array[i]
			if tmp < min {
				//fmt.Printf("index: %d i: %d tmp: %d min: %d count: %d\n", index, i, tmp, min, count)
				break
			}
			sum += tmp
			count++
		}
	}
	cacheTotal[min] = sum
	return count
}