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;

using LL = long long;

int main() {
  int N, M;
  cin >> N >> M;
  vector<LL> A(N);
  for(auto& i: A) cin >> i;

  sort(begin(A), end(A));
  vector<LL> S = {0};
  for(auto i: A) S.emplace_back(S.back() + i);

//   for(auto i: A) cout << " " <<i;cout<<endl;
//   for(auto i: S) cout << " " <<i;cout<<endl;

  auto sum = [&](auto l, auto r) {return max(0ll, S[r + 1] - S[l]);};

  auto score = [&](auto x) {
    LL s = 0;
    auto m = M;
//     cout<<"*"<<x<<endl;
    for(int i = N - 1; 0 <= i; --i) if(m) {
      int n = lower_bound(begin(A), end(A), x - A[i]) - begin(A);
      if(i < n) continue;
      auto d = 2 * (i - n) + 1;
      if(d <= m) {
        m -= d;
        s += 2 * sum(n, i - 1) + 2 * (i - n) * A[i] + 2 * A[i];
//         cout<<"all: "<<n<<","<<i<<"; "<<s<<endl;
      } else {
        s += 2 * A[i];
        if(!--m) continue;
//         cout<<"over: "<<m<<", "<<n<<","<<i<<"; "<<s<<endl;
        d = m / 2;
        s += 2 * sum(i - d, i - 1) + 2 * d * A[i];
        if(!(m -= 2 * d)) continue;
//         cout<<"over: "<<m<<", "<<n<<","<<i<<"; "<<s<<"   "<<d<<endl;
        s += A[i] + A[i - d - 1];
        --m;
//         cout<<"over: "<<m<<", "<<n<<","<<i<<"; "<<s<<endl;
      }
    }
    return s;
  };

  auto ok = [&](auto x) {
    auto m = 0;
//     cout<<"*"<<x<<endl;
    for(auto i = 0; i < N; ++i) {
      int n = lower_bound(begin(A), end(A), x - A[i]) - begin(A);
      if(n <= i) m += 2 * (i - n) + 1;
    }
//     cout<<x<<","<<m<<endl;
    return M <= m;
  };

  LL l = 0, h = 100000+1;
  while(h - l != 1) {
    auto m = (l + h) / 2;
    if(ok(m)) l = m;
    else      h = m;
  }
  cout << score(l) << endl;
}