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: Accepted)

package main

import (
	"fmt"
	"sort"
)

func main() {
	var n, m int
	var score int64
	var power []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.Ints(power)

	min := findByBTree(power, m, numOfElem)

	score, minNum, count := calcTotal(power, min)

	if m < count {
		score -= int64(minNum * (count - m))
	}

	fmt.Print(score)
}

func calcTotal(array []int, min int) (int64, int, int) {
	l := len(array)
	total := int64(0)
	minNum := 200000
	count := 0
	cache := []int64{0}
	cacheBound := make(map[int]int)
	sum := 0
	for _, a := range array {
		cache = append(cache, int64(a+sum))
		sum += a
	}

	max := cache[len(cache) - 1]
	for _, a := range array {
		bound := 0
		if val, ok := cacheBound[a]; ok {
			bound = val
		} else {
			bound = findLowerBound(array, min - a)
			cacheBound[a] = bound
		}
		if bound != -1 {
			count += l - bound
			total += max - cache[bound] + int64(a * (l - bound))
			if minNum > a + array[bound] {
				minNum = a + array[bound]
			}
		}
	}
	return total, minNum, count
}

func findByBTree(array []int, target int, check func(array []int, min int, target int) int) int {
	left := array[0]
	right := array[len(array)-1] * 2
	mid := 0
	count := 0
	var countCache []int
	var midCache []int
	for left <= right {
		mid = (left + right) / 2
		count = check(array, mid, target)
		if count == target {
			return mid
		} else if count > target {
			left = mid + 1
		} else if count < target {
			right = mid - 1
		}
		countCache = append(countCache, count)
		midCache = append(midCache, mid)
	}

	if count >= target {
		return mid
	}
	for i := len(countCache) - 1; i >= 0; i-- {
		if countCache[i] > target {
			return midCache[i]
		}
	}
	return -1
}

func numOfElem(array []int, min int, target int) int {
	len := len(array)
	count := 0
	cacheBound := make(map[int]int)
	for i := len - 1; i >= 0; i-- {
		if target < count {
			break
		}
		index := 0
		if val, ok := cacheBound[min - array[i]]; ok {
			index = val
		} else {
			index = findLowerBound(array, min-array[i])
			cacheBound[min-array[i]] = index
		}
		if index != -1 {
			count += len - index
		}
	}

	return count
}

func findLowerBound(array []int, target int) int {
	l := len(array)
	left := 0
	right := l - 1
	mid := -1
	tmp := 0
	var tmpCache []int
	var midCache []int
	for left <= right {
		mid = (left + right) / 2
		tmp = array[mid]
		if tmp == target {
			for mid > 1 && array[mid] == array[mid-1] {
				mid--
			}
			break
		} else if tmp < target {
			left = mid + 1
		} else if tmp > target {
			right = mid - 1
		}
		tmpCache = append(tmpCache, tmp)
		midCache = append(midCache, mid)
		if left > right {
			if left == l {
				mid = -1
			}
			break
		}
	}

	if tmp >= target {
		return mid
	}
	for i := len(tmpCache) - 1; i >= 0; i-- {
		if tmpCache[i] > target {
			return midCache[i]
		}
	}
	return -1
}