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<bits/stdc++.h>
using namespace std;
using ll=long long;
const int MOD=(int)1e9+7;
const int INF=(int)1e9;
const ll LINF=(ll)1e18;

int main(){
    ll n,m;
    cin>>n>>m;
    ll a[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    sort(a,a+n);
    ll sum[n+1];
    sum[n]=0;
    for(int i=n-1;i>=0;i--){
        sum[i]=sum[i+1]+a[i];
    }
    int ok=2,ng=200001;
    ll hold[200001];
    while(ng-ok>1){
        int mid=(ok+ng)/2;
        ll t=0;
        for(int i=0;i<n;i++){
            t+=n-(lower_bound(a,a+n,mid-a[i])-a);
        }
        hold[mid]=t;
        if(t>=m)ok=mid;
        else ng=mid;
    }
    ll ans=0;
    for(int i=0;i<n;i++){
        int tmp=lower_bound(a,a+n,ok-a[i])-a;
        ans+=sum[tmp]+(ll)(n-tmp)*a[i];
    }
    ans-=(hold[ok]-m)*ok;
    cout<<ans<<endl;
}