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 (Clang 3.8.0) (judge verdict: Accepted)

#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

long long count(long long n, vector<long long>& vct, long long s) {
    long long cnt = 0;
    for (long long i = 0; i < n; i++) {
        if (vct[i] * 2 >= s) {
            long long l = i;
            long long r = n - 1;
            while (l < r) {
                long long mid = (l + r) / 2 + 1;
                if (vct[i] + vct[mid] >= s) {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            cnt += 1 + 2 * (l - i);
        }
    }
    return cnt;
}

long long calculate(long long n, long long m, vector<long long>& vct, vector<long long>& svct, long long s) {
    long long cnt = 0;
    long long val = 0;
    for (long long i = 0; i < n; i++) {
        if (vct[i] * 2 >= s) {
            long long l = i;
            long long r = n - 1;
            while (l < r) {
                long long mid = (l + r) / 2 + 1;
                if (vct[i] + vct[mid] >= s) {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            val += vct[i] * 2 + 2 * (svct[l + 1] - svct[i + 1] + vct[i] * (l - i));
            cnt += 1 + 2 * (l - i);
        }
    }
    val -= s * (cnt - m);
    return val;
}

int main() {
    long long n, m;
    scanf("%lld %lld", &n, &m);

    vector<long long> vct = vector<long long>(n);
    for (long long i = 0; i < n; i++) {
        scanf("%lld", &vct[i]);
    }
    sort(vct.rbegin(), vct.rend());

    vector<long long> svct = vector<long long>(n + 1);
    for (long long i = 0; i < n; i++) {
        svct[i + 1] = svct[i] + vct[i];
    }

    long long l = 1;
    long long r = 2000000;
    while (l < r) {
        long long mid = (l + r) / 2 + 1;
        long long cnt = count(n, vct, mid);
        if (cnt >= m) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }

    long long res = calculate(n, m, vct, svct, l);
    printf("%lld\n", res);
}