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<iostream>
#include<math.h>
#include<algorithm>
#include<stdint.h>
#include<vector>
#include<deque>
#include<stack>
#include<functional>
#include<string>
#include<cstring>
#include<time.h>
#include<array>
#include<iomanip>
#include<list>
#include<set>
#include<map>
#include<random>
#include<unordered_map>
#include<unordered_set>
#include<bitset>
#include <queue>


using namespace std;
using ll = long long;
using ull = unsigned long long;
using uint = unsigned int;
using ldouble = long double;


//BDD,ZDD,kdtree, bdtree,bicalc, bolonoy, doloney, tree, chinesemod,segmenttree,daikusutora, saidairyuu, 2bugurahu, heirokenshutu, topologicalsort, kyourenketuseibun

#define REP(i,a,b) for(int i = a; i < b; ++i)
#define PRI(s) cout << s << endl
#define PRIY PRI("Yes")
#define PRIN PRI("No")


int N,M;
vector<ll> A;
vector<ll> S;

pair<ll,ll> f(ll x) {
	ll ans = 0, tot = 0;
	for (ll a : A) {
		auto itr = lower_bound(A.begin(), A.end(), x - a);
		ll d = distance(A.begin(), itr);
		ll c = A.size() - d;
		tot += S[A.size()] - S[d];
		tot += a * c;
		ans += c;
	}
	return pair<ll, ll>(tot, ans);
}

ll bins(ll m) {
	ll l = 0;
	ll r = 10000000;
	while (r - l > 1) {
		ll mid = (r + l) / 2;
		ll tmp = f(mid).second;
		if (tmp >= m) l = mid;
		else r = mid;
	}
	return l;
}

ll solve(ll L) {
	auto p = f(L);
	ll ans = p.first;
	ans -= (p.second - M) * L;
	return ans;
}

int main() {
	/*
	cin >> N >> M;
	A.resize(N);
	REP(i, 0, N) cin >> A[i];
	sort(A.begin(), A.end());
	S.resize(N+1);
	S[0] = 0;
	REP(i, 1, N + 1)S[i] = S[i - 1] + A[i - 1];
	PRI(solve(bins(M)));*/
	cin >> N >> M;
	A.resize(N);
	REP(i, 0, N)cin >> A[i];
	sort(A.begin(), A.end());
	S.resize(N + 1);
	S[0] = 0;
	REP(i, 0, N) S[i + 1] = S[i] + A[i];

	auto calc = [&](int x) {
		ll tot = 0, num = 0;
		REP(i, 0, N) {
			auto itr = lower_bound(A.begin(), A.end(), x - A[i]);
			int d = distance(A.begin(), itr);
			num += N - d;
			tot += S[N] - S[d];
			tot += A[i] * (ll)(N - d);
		}
		return pair<ll, ll>(tot, num);
	};
	int l = 0, r = 10000000;
	while (l + 1 < r) {
		int c = (l + r) / 2;
		if (calc(c).second >= M) l = c; else r = c;
	}
	auto p = calc(l);
	ll ans = p.first;
	ans -= (p.second - M) * l;
	PRI(ans);
	return 0;
}