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;
int a[100005];
LL sum[100005];
int main(){
    int n;
    LL m;
    scanf("%d%lld", &n, &m);
    for(int i=0;i<n;i++) scanf("%d", &a[i]);
    sort(a, a+n);
    for(int i=0;i<n;i++){
        sum[i] = a[i];
        if(i > 0) sum[i] += sum[i-1];
    }
    int lo = 2, hi = 2e5;
    LL ans = 0;
    while(lo <= hi){
        int mid = (lo+hi)/2;
        LL ways = 0, res = 0;

        for(int i=0;i<n;i++){
            int j = lower_bound(a, a+n, mid-a[i]) - a;
            int l = n-j;
            ways += l;
            if(j-1 >= 0) res += (sum[n-1] - sum[j-1]) + l*1ll*a[i];
            else res += sum[n-1] + l*1ll*a[i];
        }

        if(ways >= m){
            ans = res-mid*1ll*ways+mid*1ll*m;
            lo = mid+1;
        }
        else hi = mid-1;
    }
    printf("%lld\n", ans);
}