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;
const int maxn = 1e5 + 5;
ll arr[maxn], prefix[maxn];
ll check(int val, int n) {
    ll cnt = 0;
    for (int i = 1; i <= n; ++i) {
        int pos = lower_bound(arr + 1, arr + 1 + n, val - arr[i]) - arr;
        cnt += n + 1 - pos;
    }
    return cnt;
}
int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    int n; ll m; cin >> n >> m;
    for (int i = 1; i <= n; ++i) cin >> arr[i];
    sort(arr + 1, arr + 1 + n);
    for (int i = 1; i <= n; ++i) prefix[i] = prefix[i - 1] + arr[i];
    int left = 0, right = 2 * maxn; ll res = 0;
    while (left <= right) {
        int mid = (left + right) / 2; ll cnt;
        if ((cnt = check(mid, n)) < m)
            right = mid - 1;
        else {
            left = mid + 1, res = 0;
            for (int i = 1; i <= n; ++i) {
                int pos = lower_bound(arr + 1, arr + 1 + n, mid - arr[i]) - arr;
                res += (n + 1 - pos) * arr[i] + prefix[n] - prefix[pos - 1];
            }
            if (cnt > m) res -= (cnt - m) * mid;
        }
    }
    cout << res << '\n';
}