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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <vector>
#include <list>
#include <map>
#include <string>
#include <iostream>
#include <fstream>
#include <algorithm>
#include <cmath>

using namespace std;
typedef long long ll;
typedef long double ld;

int solve(void);
ll count_shakehands(ll X, ll& ans);

int main(void)
{
	while (solve())
	{
	}
	return 0;
}

ll N, M;
vector<ll> A, B;

int solve(void)
{
	cin >> N >> M;
	A = vector<ll>(N);

	for (auto& a :A)
		cin >> a;

	sort(A.begin(), A.end(), std::greater<ll>());
	B = A;
	for (ll i = 0; i < A.size() - 1; i++)
		B[i+1] += B[i];

	ll ng = A[0] * 3;
	ll ok = 0;

	ll ans = 0;

	while (abs(ok - ng) > 1)
	{
		ll mid = (ng + ok) / 2;
		ll cnt = count_shakehands(mid, ans);
		( cnt >= M ? ok : ng ) = mid;
	}
	
	ll cnt = count_shakehands(ok, ans);
	cout << ans - ok * (cnt - M) << endl;
		
	return 0;
}

ll count_shakehands(ll X, ll& ans)
{
	ll cnt = 0;
	ans = 0;

	for (ll i = 0; i < A.size(); i++)
	{
		if (A[i] + A[0] < X) { continue; }

		ll ok = upper_bound(
			A.begin(),
			A.end(),
			X - A[i], greater<ll>()
		) - A.begin();

		cnt += (ok);

		{
			ll sa = A[i] * ( ok );
			ll sb = B[ok - 1];
			ans += sa + sb;
		}
	}

	return cnt;
}