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 <cmath>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <stack>
#include <set>
#include <string>
#include <utility>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

const ll M = 1e9+7;
const ll INF = 1e9+7;

ll n, m, ans;
ll a[110000];

ll binary_search(ll m) {
  ll xl = 0, xr = 1e6, x;
  while (xl < xr) {
    x = (xl+xr)/2;
    ll cnt = 0;
    for (ll i = 0; i < n; i++) {
      ll l = 0, r = n, mid;  
      while (l < r) {
        mid = (l + r) / 2;
        if (a[i] + a[mid] >= x) r = mid;
        else l = mid + 1;
      }
      cnt += n-l;
    }
    
    if (cnt < m) {
      xr = x;
    } else {
      xl = x+1;
    }
  }
  return xl-1;
}

int main() {
  cin >> n >> m;

  for (ll i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a+n);
  
  ll x = binary_search(m);
  
  ll idx = n-1;
  ll sum = 0;
  ll cnt = 0;
  for (ll i = 0; i < n; i++) {
    cnt += n - idx - 1;
    if (cnt > m) {
      cnt -= n - idx - 1;
      for (ll j = n-1; j >= n-m+cnt; j--) {
        ans += a[i] + a[j];
      }
    }
    while (idx >= 0 and a[i] + a[idx] >= x and cnt + n - idx - 1 < m) {
      sum += a[idx];
      idx--;
    }
    ans += a[i] * (n-idx-1) + sum;
  }
  
  cout << ans << endl;
  return 0;
}