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;
using Graph = vector<vector<int>>;
using WGraph = vector<vector<pair<int, ll>>>;

int main() {
    int n, m;
    cin >> n >> m;
    vector<ll> a(n);
    for (int i=0; i<n; i++) {
        cin >> a.at(i);
    }
    vector<ll> cnt(2e5+1, 0), sum(2e5+1, 0);
    for (int i=0; i<n; i++) {
        cnt.at(a.at(i))++;
        sum.at(a.at(i)) += a.at(i);
    }
    for (int i=2e5-1; i>=0; i--) {
        cnt.at(i) += cnt.at(i+1);
        sum.at(i) += sum.at(i+1);
    }
    ll ok = -1;
    ll ng = 2e5+1;
    while (abs(ok-ng) > 1) {
        int mid = (ok+ng)/2;
        ll tmp = 0;
        for (int i=0; i<n; i++) {
            tmp += (mid - a.at(i) >= 0) ? cnt.at(mid-a.at(i)) : cnt.at(0);
        }
        if (tmp >= m) ok = mid;
        else ng = mid;
    }
    ll ans = 0;
    for (int i=0; i<n; i++) {
        ans += a.at(i) * ((ok - a.at(i) >= 0) ? cnt.at(ok-a.at(i)) : cnt.at(0)) + ((ok - a.at(i) >= 0) ? sum.at(ok-a.at(i)) : sum.at(0));
    }
    ll tmp = 0;
    for (int i=0; i<n; i++) {
        tmp += (ok - a.at(i) >= 0) ? cnt.at(ok-a.at(i)) : cnt.at(0);
    }
    if (tmp > m) {
        ans -= ok * (tmp - (ll)m);
    }
    cout << ans << endl;
    return 0;
}