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: Runtime Error)

#include <iostream>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <string>
#include <sstream>
#include <algorithm>
#include <random>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>
#include <cassert>
#include <climits>
#include <bitset>
#include <functional>
#include <iomanip>

#define FOR_LT(i, beg, end) for (int i = beg; i < end; i++)
#define FOR_LE(i, beg, end) for (int i = beg; i <= end; i++)
#define FOR_DW(i, beg, end) for (int i = beg; end <= i; i--)

using namespace std;

struct L {
	int64_t i;
	int64_t j;
	int64_t s;
};
int64_t lval(const L& l, const vector<int64_t>& as) {
	return  as[l.i] + as[l.j];
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout << fixed << setprecision(20);

	int64_t n, m; cin >> n >> m;
	vector<int64_t> as(n);
	FOR_LT(i, 0, n) {
		cin >> as[i];
	}
	sort(as.begin(), as.end(), greater<int64_t>());

	vector<int64_t> ss(n);
	{
		int64_t s = 0;
		FOR_LT(i, 0, n) {
			s += as[i];
			ss[i] = s;
		}
	}

	vector<L> ls(n);
	FOR_LT(i, 0, n) {
		ls[i].i = i;
		ls[i].j = -1;
	}

	auto cmp = [](const L& lhs, const L& rhs) {
		return lhs.s > rhs.s;
	};
	priority_queue<L, vector<L>, decltype(cmp)> queue(cmp);
	int ci = 0;
	int cj = n - 1;
	{
		int64_t mm = m;
		int i = 0;
		while (mm != 0) {
			ls[i].j = min(mm, n) - 1;
			ls[i].s = lval(ls[i], as);
			mm -= min(mm, n);
			if (mm == 0) {
				ci = i;
				break;
			}
			queue.push(ls[i]);
			i++;
		}
	}

	if (queue.size() == 0) {
		queue.push(ls[0]);
		ci++;
	}
	while (0 <= cj && ci < n) {
		L ql = queue.top();
		if (ls[ci].j == cj) {
			ls[ci].s = lval(ls[ci], as);
			//queue.push(ls[ci]);
			ci++;
			continue;
		}

		int64_t v = as[ci] + as[cj];
		if (v > ql.s) {
			ls[ci].j++;
			ls[ci].s = lval(ls[ci], as);
			ls[ql.i].j--;
			ls[ql.i].s = lval(ls[ql.i], as);
			//queue.pop();
			//queue.push(ls[ql.i]);
		}
		else {
			cj--;
		}
	}

	int64_t ans = 0;
	FOR_LT(i, 0, n) {
		auto& l = ls[i];
		if (l.j < 0) continue;
		ans += ss[l.j] + as[i] * (l.j + 1LL);
	}
	cout << ans << endl;

	return 0;
}