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;
typedef long long ll;
#define rep(k,i,n) for(ll i=k;i<n;++i)
ll N,M;
vector<ll> A;
vector<ll> idx;
ll sum;
bool f(ll mid){
    sum=0;
    rep(0,i,N){
        idx[i]=lower_bound(A.begin(),A.end(),mid-A[i])-A.begin();//左手をiとしてmid1より少ない握手した時の組み合わせの数。
        sum+=N-idx[i];//sum=(すべての組み合わせの数)
    }
    return sum>=M;
}
int main(void){
    // Your code here!
    cin>>N>>M;
    A.resize(N);
    idx.resize(N);
    rep(0,i,N)cin>>A[i];
    sort(A.begin(),A.end());
    ll left=-1,right=1e6;
    while(left+1<right){
        ll mid=(left+right)/2;
        if(f(mid))left=mid;
        else right=mid;
    }
    f(left);
    ll ans=0;
    rep(0,i,N)ans+=(N-idx[i])*A[i]*2;
    ans-=(sum-M)*left;
    cout<<ans<<endl;
}