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 <algorithm>
#include <cassert>
#include <cmath>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define DEBUG 1
constexpr int kMod = 1000000007;
typedef long long LL;

int Count(const std::vector<LL>& A, LL v) {
  int N = A.size();
  int count = 0;
  for (const LL Ax : A) {
    LL Ay = v - Ax;
    int idx = std::lower_bound(A.begin(), A.end(), Ay) - A.begin();
    count += N - idx;
  }
  return count;
}

LL Sum(const std::vector<LL>& A, LL v) {
  int N = A.size();
  std::vector<LL> As = A;
  for (int i = N - 2; i >= 0; --i) As[i] += As[i+1];

  LL sum = 0;
  for (const LL Ax : A) {
    LL Ay = v - Ax;
    int idx = std::lower_bound(A.begin(), A.end(), Ay) - A.begin();
    LL cnt = N - idx;
    if (cnt == 0) continue;

    sum += Ax * cnt;
    sum += As[idx];
  }
  return sum;
}

int main() {
  int N, M; std::cin >> N >> M;
  std::vector<LL> A(N);
  for (int i = 0; i < N; ++i) std::cin >> A[i];
  std::sort(A.begin(), A.end());

  int l = 0, r = 2 * A[N-1];
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (Count(A, mid) >= M) l = mid;
    else                    r = mid;
  }

  LL sum = Sum(A, l) - static_cast<LL>(l * (Count(A, l) - M));
  std::cout << sum << std::endl;
}