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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <algorithm>
#include <set>
#include <iostream>
#include <cassert>
#include <queue>
using namespace std;


int main() {
    int N = 0;
    long long M = 0;
    cin >> N >> M;
    vector<int> A;
    int val = 0;
    for (int i = 0; i < N; ++i) {
        cin >> val;
        A.push_back(val);
    }
    sort(A.begin(), A.end());
    int low = 0, high = 200001;
    //找到最小的 k, 使得 happiness > k 的 组合数 < M.
    while (low < high) {
        int mid = (low + high) / 2;
        long long cnt = 0;
        int i = 0, j = N-1;
        while (i < N) {
            while (j >= 0 && A[i] + A[j] > mid) j--;
            cnt += N - j - 1;
            i++;
        }
        // cnt 与 mid 成反比
        if (cnt >= M) {
            low = mid + 1;
        } else {
            high = mid;
        }
    }
    int k = low;
    long long ans = 0;
    int i = 0, j = N-1;
    long long pre = 0;
    long long cnt = 0;
    while (i < N) {
        while (j >= 0 && A[i] + A[j] > k) {
            pre += A[j--];
        }
        long long t = N-j-1;
        cnt += t;
        ans += pre + t * A[i];
        i++;
    }
    ans += (M - cnt) * k;
    cout << ans << endl;

    return 0;
}