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>
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
#define yn {puts("Yes");}else{puts("No");}
#define MAX_N 200005

int main() {
    ll n, k;
    cin >> n >> k;

    ll a[n];
    rep(i,n)cin >> a[i];
    sort(a,a+n);

    ll ans = 0;

    ll l = 0, r = 1001001;

    while(l + 1 < r){
        ll m = (l + r) / 2;
        ll cnt = 0;
        rep(i,n){
            auto ite = lower_bound(a, a+n, m - a[i]);
            cnt += a + n - ite;
        }

        if(cnt >= k){
            l = m;
        }else{
            r = m;
        }
    }

    ll b[n];
    rep(i,n)b[i] = a[i];
    sort(b,b+n,greater<ll>());
    ll sum[n+1] = {};
    srep(i,1,n+1){
        sum[i] = sum[i-1] + b[i-1];
    }


    ll cc = 0;
    rep(i,n){
        auto ite = lower_bound(a, a+n, l - a[i]);
        ll cnt = a + n - ite;
        cc += cnt;
        // cout << i << ' ' << cnt << endl;
        ans += cnt * a[i];
        ans += sum[cnt];
    }

    if(cc > k){
        ans -= (cc - k) * l;
    }

    cout << ans << endl;
    return 0;
}