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;

// def
#define debug(x) cout << #x << ": " << x << endl
#define out(x) cout << x << endl
#define repeat(i, a, b) for (int i = (a); i < (b); i++)
#define revrepeat(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define revrep(i, n) for (int i = (n)-1; i >= 0; i--)
#define all(x) (x).begin(), (x).end()
typedef long long ll;
typedef pair<int, int> P;
typedef vector<vector<int>> Graph;
const int M = 1e9 + 7;
const ll INF = 1e9;
const int MAX_V = 100000 + 5;
const int MAX_N = 1e5 + 5;

ll n, m;
vector<ll> a;
bool valid(ll x) {
    ll cnt = 0;
    rep(i, n) {
        auto iter = lower_bound(all(a), x - a[i]);
        cnt += a.end() - iter;
    }
    if (cnt >= m) {
        return true;
    } else {
        return false;
    }
}

int main() {
    cin >> n >> m;
    a.resize(n);
    rep(i, n) cin >> a[i];
    sort(all(a));
    vector<ll> cusum(n, 0);
    rep(i, n) {
        if (i == 0) cusum[0] = a[0];
        cusum[i] = cusum[i - 1] + a[i];
    }

    ll lb = 0, ub = 1e10;
    while (ub - lb > 1) {
        ll mid = (lb + ub) / 2;
        if (valid(mid)) {
            lb = mid;
        } else {
            ub = mid;
        }
    }

    ll cnt = 0, ans = 0;
    rep(i, n) {
        auto iter = lower_bound(all(a), lb - a[i]);
        cnt = a.end() - iter;
        ans += cusum[a.end() - a.begin() - 1] - cusum[iter - a.begin() - 1];
        ans += cnt * a[i];
    }
    out(ans);
}