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 <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <functional>
#include <climits>
#include <cstring>
#include <cmath>
#include <numeric>

using namespace std;

#define int long long
#define rep(i, n) for (int i=0;i<(int)n;++i)
#define rep1(i, n) for (int i=1;i<=(int)n;++i)
#define rrep(i, n) for (int i=(int)n-1;i>=0;--i)
#define rrep1(i, n) for (int i=n;i>=1;--i)
#define range(i, l, r) for (int i=l;i<(int)r;++i)
#define rrange(i, l, r) for (int i=(int)r-1;i>=(int)l;--i)
#define unless(a) if(!(a))
#define all(a) begin(a),end(a)
#define fst first
#define scd second
#define PB emplace_back
#define PPB pop_back

using vi=vector<int>;
using pii=pair<int, int>;

bool chmin(int&a,int b){return a>b?(a=b,true):false;}
bool chmax(int&a,int b){return a<b?(a=b,true):false;}
int read(){int a;scanf("%lld",&a);return a;}
constexpr int TEN(int n){return n==0?1:10*TEN(n-1);}

const int inf = TEN(9) + 10;

int N, M;
vi A;
int rui[100010];

int calc(int x)
{
	int res = 0;
	
	rep(i, N) {
		res += upper_bound(all(A), x - A[i], greater<int>()) - begin(A);
	}

	return res;
}

signed main()
{
	cin >> N >> M;

	A.resize(N);

	rep(i, N) cin >> A[i];
	sort(all(A), greater<int>());

	int ok = 0, ng = inf;

	while (abs(ok - ng) > 1) {
		int mid = (ok + ng) / 2;

		int tmp = calc(mid);

		if (tmp >= M) {
			ok = mid;
		} else {
			ng = mid;
		}
	}

	int counter = calc(ok);

	int ans = 0;

	rep(i, N) {
		rui[i + 1] += rui[i] + A[i];
	}

	rep(i, N) {
		int pos = upper_bound(all(A), ok - A[i], greater<int>()) - begin(A);

		ans += rui[pos] + A[i] * pos;
	}

	ans -= (counter - M) * ok;

	cout << ans << endl;
}