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 Python (3.4.3) (judge verdict: Accepted)

import sys

sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]

def main():
    def ok(c):
        cnt = 0
        i = 0
        for a in aa[::-1]:
            while i < n and a + aa[i] >= c: i += 1
            cnt += i
            if cnt >= m: return True
        return False

    n, m = MI()
    aa = LI()
    aa.sort(reverse=True)
    # 握手の回数がm以上になるような、1回の握手の幸福度の最低値を求める
    l = aa[-1] * 2 - 1
    r = aa[0] * 2 + 1
    while l + 1 < r:
        c = (l + r) // 2
        if ok(c):
            l = c
        else:
            r = c
    min_once = l
    # 幸福度がmin_once以上になる握手の総和を求める
    ans = 0
    i = 0
    s = 0
    cnt = 0
    for a in aa[::-1]:
        while i < n and a + aa[i] >= min_once:
            s += aa[i]
            i += 1
        ans += s + i * a
        cnt += i
    # 握手の回数がmを越えたときはその分の幸福度を引く
    if cnt - m:
        ans -= min_once * (cnt - m)
    print(ans)

main()