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;
typedef long long ll;
#define rep(i,n) for(ll i=0;i<(int)n;i++)
#define rep1(i,n) for(int i=0;i<=(int)n;i++)
int main(void){
    // Your code here!
    int n,m;cin>>n>>m;
    vector<ll> a(n);
    rep(i,n) cin>>a[i];
    sort(a.begin(),a.end());
    //reverse(a.begin(),a.end());
    vector<ll> b(n+1,0);b[n-1]=a[n-1];
    rep1(i,n-1)b[n-i-1]=b[n-i]+a[n-1-i];
    ll sum=0,left=0,right=a[n-1]*2+1,ans=0;
    while(right-left>1){
        ll mid=(right+left)/2;
        sum=0;
        rep(i,n){
            ll buf=mid-a[i];
            auto it=lower_bound(a.begin(),a.end(),buf);
            ll k=it-a.begin();
            //cout<<n-k<<endl;
            sum+=n-k;
        }
        if(sum<=m)right=mid;
        else if(sum>m)left=mid;
    }
    //cout<<right<<endl;
    rep(i,n){
        ll buf=right-a[i];
        auto it=lower_bound(a.begin(),a.end(),buf);
        ll k=it-a.begin();
        ans+=(n-k)*a[i]+b[k];
        m-=n-k;
    }
    ans+=m*left;
    if(n*n==m){
        ans=(n+1)*b[0];
    }
    cout<<ans<<endl;
    return 0;
}