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 (1.6) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
)

func out(x ...interface{}) {
	fmt.Println(x...)
}

var sc = bufio.NewScanner(os.Stdin)

func getInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func getString() string {
	sc.Scan()
	return sc.Text()
}

// min, max, asub, absなど基本関数
func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func asub(a, b int) int {
	if a > b {
		return a - b
	}
	return b - a
}

func abs(a int) int {
	if a >= 0 {
		return a
	}
	return -a
}

func lowerBound(a []int, x int) int {
	l := 0
	r := len(a) - 1
	for l <= r {
		m := (l + r) / 2
		if a[m] >= x {
			r = m - 1
		} else {
			l = m + 1
		}
	}
	return l
}

func calc(a, s []int, x int) (int, int) {
	tot := 0
	num := 0
	n := len(a)
	for i := 0; i < n; i++ {
		j := lowerBound(a, x-a[i])
		num += n - j
		tot += s[n] - s[j]
		tot += a[i] * (n - j)
	}
	return tot, num
}

func main() {
	sc.Split(bufio.ScanWords)
	N, M := getInt(), getInt()
	a := make([]int, N)
	for i := 0; i < N; i++ {
		a[i] = getInt()
	}
	sort.Ints(a)
	b := make([]int, N+1)
	for i := 0; i < N; i++ {
		b[i+1] = b[i] + a[i]
	}

	l := 0
	r := 200005
	for l+1 < r {
		c := (l + r) >> 1
		_, num := calc(a, b, c)
		if num >= M {
			l = c
		} else {
			r = c
		}
	}
	ans, num := calc(a, b, l)
	ans -= (num - M) * l
	out(ans)
}