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)

from bisect import bisect_left
import numpy


def main():
    def isOK(score):
        count = 0
        for a in reversed(A):
            border_score = score - a
            index = bleft(A, border_score)
            count += N - index
            if count >= M:
                return True
        return False

    N, M = map(int, input().split())
    A = sorted(map(int, input().split()))
    ans = 0

    ng = A[-1] + A[-1] + 1
    ok = A[0] + A[0]
    bleft = bisect_left

    while(abs(ok-ng) > 1):
        #print(ok, ng)
        mid = (ok+ng) // 2
        f = isOK(mid)
        if f:
            ok = mid
        else:
            ng = mid
        #print(f)
        #print(ok, ng)
    C = [0,] + A
    C = numpy.cumsum(C[::-1])
    #print(C)
    min_score = float('inf')
    count = 0
    for a in reversed(A):
        index = bleft(A, ok-a)
        if index < N:
            min_score = min(a+A[index], min_score)
            ans += C[N-index-1] + (N-index) * a
            count += N-index

    ans -= min_score * (count-M)
    print(ans)


if __name__ == '__main__':
    main()