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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <functional>
#include <queue>
#include <unordered_set>
#include <climits>

using namespace std;
typedef long long ll;

bool IsMore(ll A[],ll S[], int N, int M, int K, ll& sum){
	int count = 0;
	sum = 0;
	for(int i = 0; i < N; i++){
		int left = A[i];
		int min = 0;
		int max = N;
		while(max - min > 0){
			int mid = (max+min)/2;
			//cout << min << " " << mid << " " << max << endl;
			if(mid == N){
				min = mid;
				max = mid;
				break;
			}
			else if(left + A[mid] >= K){
				max = mid;
			}
			else{
				min = mid + 1;
			}
		}
		count += N - max;
		sum += A[i] * (N-max) + S[max];
	}

	sum -= (count-M)*K;
	
	return count >= M;
}

int main()
{
	int N, M;
	ll A[100000];
	cin >> N >> M;
	for(int i = 0; i < N; i++){
		cin >> A[i];
	}
	sort(A, A+N);
	
	ll S[100000];
	S[N-1] = A[N-1];
	for(int i = N-1; i > 0; i--){
		S[i-1] = S[i] + A[i-1];
	}
	
	int min = 0;
	int max = 200000;
	ll sum;
	// 幸福度がX以上になるような組み合わせがM以上あるような最小のXを二分探索
	while(max - min > 0){
		int mid = (max+min+1)/2;
		if(IsMore(A, S, N, M, mid, sum)){
			min = mid;
		}else{
			max = mid - 1;
		}
	}
	
	IsMore(A, S, N, M, min, sum);
	
	cout << sum << endl;
	
	return 0;
}