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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
	int N;
	long long M;
	cin >> N >>M;
	vector<int> A(N);
	for (int n = 0;n < N;++n) {
		cin >> A[n];
	}
	sort(A.begin(), A.end());
	int l = 0;
	int r = 200005;
	vector<long long>S(N + 1);
	for (int n = 0;n<N;++n) {
		S[n + 1] = (long long)A[n] + S[n];
	}

	auto calc = [&](int x) {
		ll tot = 0, num = 0;
		rep(i, N) {
			int j = lower_bound(A.begin(), A.end(), x - A[i]) - A.begin();
			num += N - j;
			tot += S[N] - S[j];
			tot += A[i] * ll(N - j);
		}
		return make_pair(tot, num);
	};

	while (l + 1 < r) {
		long long num = 0;
		int c = (l+r) / 2;
		/*for (int n = 0;n < N;++n) {
			int j = lower_bound(A.begin(), A.end(), c - A[n]) - A.begin(); 
			num += (long long)(N - j);
		}
		if (num >= M) {
			l = c;
		}
		else {
			r = c;
		}*/
		if (calc(c).second >= M) l = c; else r = c;
	}
	long long ans = 0;
	long long num = 0; 
	for (int n = 0; n < N; ++n) {
		int j = lower_bound(A.begin(), A.end(), l - A[n]) - A.begin();
		num += (long long)(N - j);
		ans += (long long)(S[N] - S[j]);
		ans += (long long)(A[n] * (N - j));
	}

	ans -= (long long)((num-M)*l);
	cout << ans <<endl;
	/*auto p = calc(l);
	ll ans = p.first;
	ans -= (p.second - M)*l;
	cout << ans << endl;*/
	return 0;
}