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;

#define N_TIMES(i, n)     for ( uint64_t i = 0;     i <  n; ++i )
#define N_TIMES_REV(i, n) for (  int64_t i = n - 1; i >= 0; --i )

bool judge(
  uint64_t N, uint64_t M,
  const vector<uint64_t> &A,
  const vector<uint64_t> &B,
  uint64_t D
)
{
  uint64_t C = 0;

  N_TIMES(n, N) {
    uint64_t diff = D - A[n];
    if (D < A[n]) { diff = 0; }

    auto itr = lower_bound(A.begin(), A.end(), diff, greater_equal<uint64_t>());
    uint64_t d = distance(A.begin(), itr);

    C += d;
  }

  return C >= M;
}

uint64_t calculate(
  uint64_t N, uint64_t M,
  const vector<uint64_t> &A,
  const vector<uint64_t> &B,
  uint64_t D
)
{
  uint64_t T = 0, m = 0;

  N_TIMES(n, N) {
    uint64_t diff = D - A[n];
    if (D < A[n]) { diff = 0; }

    auto itr = lower_bound(A.begin(), A.end(), diff, greater_equal<uint64_t>());
    uint64_t d = distance(A.begin(), itr);

    T += B[d] + (A[n] * d);
    m += d;
  }
  
  T -= (m - M) * D;

  return T;
}

int main()
{
  uint64_t N, M;
  cin >> N >> M;

  vector<uint64_t> A(N);
  N_TIMES(n, N) { cin >> A[n]; }

  sort(A.begin(), A.end(), greater_equal<uint64_t>());

  vector<uint64_t> B(N + 1, 0);
  N_TIMES(n, N) { B[n + 1] = A[n] + B[n]; }

  uint64_t hi = 1e+5 + 10, lo = 0;

  while (hi - lo > 1) {
    uint64_t mid = (hi + lo) >> 1;

    if (judge(N, M, A, B, mid)) {
      lo = mid;
    } else {
      hi = mid;
    }
  }

  if (judge(N, M, A, B, hi)) {
    cout << calculate(N, M, A, B, hi) << endl;
  } else {
    cout << calculate(N, M, A, B, lo) << endl;
  }

  return 0;
}