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 <algorithm>
#include <vector>

using namespace std;
using ll = long long;
using vi = vector<int>;
using vl = vector<long long>;

#define in(v) v; cin >> v;
#define rep(i,n) for(int i=0,_i=(n);i<_i;++i)
#define all(f,c,...) (([&](decltype((c)) cccc) { return (f)(begin(cccc), end(cccc), ## __VA_ARGS__); })(c))
int main() {
  int in(N); ll in(M);

  vi A(N);
  rep(i, N) cin >> A[i];
  all(sort, A);
  vl sum(N+1, 0);
  rep(i, N) {
    sum[i+1] = sum[i] + A[i];
  }

  auto f = [&](ll mid) {
    // mid以上がいくつあるか？
    ll count = 0;
    rep(i, N) {
      count += A.end() - all(lower_bound, A, mid - A[i]);
    }
    return count < M;
  };
  // 上からM個取りたい
  // -> 上からM番目を探す
  ll ng = 0, ok = 1800000000000000000ll; // okの数字は上からM番目の数以上
  while (abs(ok - ng) > 1) {
    ll mid = (ok + ng) / 2;
    if (f(mid))
      ok = mid;
    else
      ng = mid;
  }
  ll ans = 0;
  ll count = 0;
  rep(i, N) {
    ans += sum[N] - sum[(int)(all(lower_bound, A, ok - A[i]) - A.begin())];
    ans += 1ll * (int)(A.end() - all(lower_bound, A, ok - A[i])) * A[i];
    count += A.end() - all(lower_bound, A, ok - A[i]);
  }
  ans += ng * (M - count);
  cout << ans << endl;
  return 0;
}