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: Time Limit Exceeded)

#include <algorithm>
#include <cmath>
#include <iostream>
#include <map>
#include <numeric>
#include <string>
#include <vector>

using namespace std;

struct Memo {
	long long fact = 0;
	long long fact_inv = 0;
	long long inv = 0;
};
class Combination {
private:
	int m_module = 1000000007;
	vector<struct Memo> memo;
public:
	int calculate(int, int);
	Combination(int, int);
};
int Combination::calculate(int n, int k) {
	return (memo[n].fact * (memo[k].fact_inv * memo[n - k].fact_inv) % m_module) % m_module;
}
Combination::Combination(int memo_max, int module = 1000000007) {
	m_module = module;
	memo.push_back({ 1,1,0 });
	memo.push_back({ 1,1,1 });
	for (int loop = 2; loop <= memo_max; loop++) {
		struct Memo temp;
		temp.fact = (memo[loop - 1].fact * loop) % m_module;
		temp.inv = m_module - (memo[m_module % loop].inv * int(m_module / loop)) % m_module;
		temp.fact_inv = (memo[loop - 1].fact_inv * temp.inv) % m_module;
		memo.push_back(temp);
	}
}

void A_Strings(void) {

	string S, T;
	cin >> S >> T;
	cout << T << S << flush;
	return;
}

void B_Greedy_Takahashi(void) {

	long long A, B, K;
	cin >> A >> B >> K;

	if (K < A) {
		A -= K;
	}
	else if (K < A + B) {
		B -= (K - A);
		A = 0;
	}
	else {
		A = 0;
		B = 0;
	}
	cout << A << " " << B << flush;
	return;
}

void C_Next_Prime(void) {
	
	int X;
	cin >> X;

	for (int sample = X;; sample++) {
		for (int loop = 2; loop <= (int)pow(sample, 0.5); loop++) {
			if (sample % loop == 0) goto nextSample;
		}
		cout << sample << flush;
		return;
	nextSample:
		continue;
	}
}

void D_Prediction_and_Restriction(void) {

	int N, K;
	int point[3];
	cin >> N >> K >> point[0] >> point[1] >> point[2];
	vector<int> T(N);
	for (int& it : T) {
		char C;
		cin >> C;
		switch (C) {
		case 'r':
			it = 2;
			break;
		case 's':
			it = 0;
			break;
		case 'p':
			it = 1;
			break;
		}
	}

	for (int rem = 0; rem < K; rem++) {
		for (int loop = K + rem; loop < N; loop += K) {
			if (T[loop - K] == T[loop]) {
				T[loop] = -1;
			}
		}
	}

	int sum = 0;
	for (int it : T) if (it != -1) sum += point[it];
	cout << sum << flush;
	return;
}

void E_Handshake(void) {

	int N;
	long long M;
	cin >> N >> M;

	vector<int> A(N);
	for (int& it : A)cin >> it;
	sort(A.begin(), A.end(), greater<int>());

	vector<long long> num(A[0] * 2 + 1);

	for (int loop1 = 0; loop1 < N; loop1++) {
		for (int loop2 = loop1; loop2 < N; loop2++) {
			if (loop1 == loop2) num[A[loop1] + A[loop2]]++;
			else				num[A[loop1] + A[loop2]] += 2;
		}
	}

	long long index = 0;
	long long sum = 0;
	for (int loop = A[0] * 2; loop >= 0; loop--) {

		if (num[loop] >= M - index) {
			sum += loop * (M - index);
			break;
		}
		else {
			sum += loop * num[loop];
			index += num[loop];
		}
	}

	/*
	long long flag = 0;
	vector<int> point((long long)N * (long long)N);
	for (int x = 0; x < N; x++) {
		for (int y = 0; y < N; y++) {
			point[flag] = A[x] + A[y];
			flag++;
		}
	}
	sort(point.begin(), point.end(), greater<int>());
	*/
	/*
	int x = 0;
	int y = 0;
	long long flag = 0;
	long long sum = 0;
	while (flag < M) {

		if (x < y && flag < M - 1) {
			sum += (A[x] + A[y]) * 2;
			flag += 2;
		}
		else {
			sum += A[x] + A[y];
			flag++;
		}

		if (x > 0 && y < N - 1 && A[x - 1] + A[y + 1] > A[x + 1] + A[y]) {
			x--;
			y++;
		}
		else if (y < N - 1 && (A[x] + A[y + 1] > A[x + 1] + A[y] || x == y)) {
			y++;
		}
		else if (x < N - 1) {
			x++;
		}
	}
	*/
	cout << sum << flush;
	return;
}

int main(void) {

	//A_Strings();
	//B_Greedy_Takahashi();
	//C_Next_Prime();
	//D_Prediction_and_Restriction();
	E_Handshake();
	return 0;
}