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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import sys,collections as cl,bisect as bs
sys.setrecursionlimit(100000)
input = sys.stdin.readline
mod = 10**9+7
Max = sys.maxsize
def l(): #intのlist
    return list(map(int,input().split()))
def m(): #複数文字
    return map(int,input().split())
def onem(): #Nとかの取得
    return int(input())
def s(x): #圧縮
    a = []
    if len(x) == 0:
        return []
    aa = x[0]
    su = 1
    for i in range(len(x)-1):
        if aa != x[i+1]:
            a.append([aa,su])
            aa = x[i+1]
            su = 1
        else:
            su += 1
    a.append([aa,su])
    return a
def jo(x): #listをスペースごとに分ける
    return " ".join(map(str,x))
def max2(x): #他のときもどうように作成可能
    return max(map(max,x))
def In(x,a): #aがリスト(sorted)
    k = bs.bisect_left(a,x)
    if k != len(a) and a[k] ==  x:
        return True
    else:
        return False

def pow_k(x, n):
    ans = 1
    while n:
        if n % 2:
            ans *= x
        x *= x
        n >>= 1
    return ans






N,M = m()

a = l()

a.sort()

po = [0 for i in range(10**5+1)]
ans = 0
for i in range(N):
    po[a[i]] += 1
for i in range(10**5-1,-1,-1):
    po[i] += po[i+1]
ll = 0
rr = 10**5 * 2

while True:
    mid = (ll+rr)//2
    if mid == ll:
        co = 0
        for i in range(N):
            ui = a[i]
            if mid > ui + 10**5:
                continue
            elif ui >= mid:
                co += N
            else:
                co += po[mid-ui]
        cco = 0
        mid = rr

        for i in range(N):
            ui = a[i]
            if mid > ui + 10**5:
                continue
            elif ui >= mid:
                cco += N
            else:
                cco += po[mid-ui]
        if cco == M:
            mid = rr
        else:
            mid = ll

        break

    co = 0
    for i in range(N):
        ui = a[i]
        if mid > ui + 10**5:
            continue
        elif ui >= mid:
            co += N
        else:
            co += po[mid-ui]
    
    if co < M:
        rr = mid-1                    
    else:
        ll = mid


a.sort(reverse = True)
dp = [0]

ans -= mid * (co - M)

for i in range(N):
    dp.append(dp[-1]+a[i])

for i in range(N):
    pl = a[i]
    if pl >= mid:
        ans += pl*N + dp[-1]
        
    elif pl + 10**5 < mid:
        continue
    else:
        ans += pl*po[mid-pl] + dp[po[mid-pl]]
print(ans)