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;
#define rep(i, N) for (int i = 0; i < (int)N; ++i)
#define all(a) (a).begin(), (a).end()

ll N;
ll M;
vector<ll> A;

int main() {
  cin >> N >> M;
  A.resize(N);
  rep(i, N) cin >> A[i];

  sort(all(A));

  vector<ll> S(N + 1, 0);
  rep(i, N) S[i + 1] = S[i] + A[i];

  function<ll(ll)> num = [&](ll t) {
    ll res = 0;
    rep(i, N) res += distance(lower_bound(all(A), t - A[i]), A.end());
    return res;
  };

  ll l = 0, r = 1e6;
  while (r - l > 1) {
    ll m = (l + r) / 2;
    if (num(m) <= M) {
      r = m;
    } else {
      l = m;
    }
  }

  ll ans = 0;
  ll rest = M;
  rep(i, N) {
    int l = distance(lower_bound(all(A), r - A[i]), A.end());
    ans += l * A[i] + S[N] - S[N - l];
    rest -= l;
  }
  ans += rest * l;

  cout << ans << endl;

  return 0;
}