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 <bits/stdc++.h>
using namespace std;
#define all(x) (x).begin(),(x).end()
#define YES() printf("YES\n")
#define NO() printf("NO\n")
#define isYES(x) printf("%s\n",(x) ? "YES" : "NO")
#define Yes() printf("Yes\n")
#define No() printf("No\n")
#define isYes(x) printf("%s\n",(x) ? "Yes" : "No")
#define isIn(x,y,h,w) (x >= 0 && x < h && y >= 0 && y < w)

#define int long long
//using ll = long long;
using P = pair<int,int>;

ostream &operator<<(ostream &os,const P &p){ return os << "(" << p.first << "," << p.second << ")"; }

template<class T> T &chmin(T &a,const T &b){ return a = min(a,b); }
template<class T> T &chmax(T &a,const T &b){ return a = max(a,b); }
 
const int INF=1e+18;
const double EPS=1e-9;
const int MOD=1000000007;

const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};

int n,m,a[100010],imos[100010];

int sum,cnt;

bool isOK(int x){
	sum = cnt = 0;
	for(int i = 0;i < n;i++){
		int ind = lower_bound(a,a + n,x - a[i]) - a;
		cnt += n - ind;
		sum += imos[n] - imos[ind] + (n - ind) * a[i];
	}
	return cnt <= m;
}

signed main(){
	cin >> n >> m;
	for(int i = 0;i < n;i++) cin >> a[i];
	sort(a,a + n);
	for(int i = 0;i < n;i++) imos[i + 1] = imos[i] + a[i];
	int ok = INF,ng = 0;
	while(ok - ng > 1){
		int mid = (ok + ng) / 2;
		if(isOK(mid)) ok = mid;
		else ng = mid;
	}
	isOK(ok);
	cout << sum + ng * (m - cnt) << endl;
}