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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>
#include <vector>
#include <algorithm>
#include <map>

int N;
long long M;
std::vector<long long > A(100000);
std::vector<long long > kumisuu(200000, 0); // A_i>=x を満たすiの数
std::vector<long long > souwa(200000, 0); // A_i>=x を満たすiについてA_iの総和

long long get_possible_count(int value){ // 値がvalue以上となる組み合わせ総数を求める
  long long count = 0;
  for (int i=0; i<N; i++){
    count += kumisuu[std::max(value-A[i], (long long)0)];
  }
  return count;
}

int main(){
  scanf("%d",&N);
  scanf("%lld",&M);

  for (int i=0; i<N; i++){
    scanf("%lld",&A[i]);
    kumisuu[A[i]]++;
  }
  sort(A.begin(), A.end(), std::greater<long long>());

  for (int i=0; i<N; i++){
    souwa[i+1] = souwa[i] + A[i];
  }

  /*  累積和を求める */
  for (int i=200000; i>0; i--){
    kumisuu[i-1] += kumisuu[i];
  }

  int left = 0;
  int right = 200001;
  int mid;
  long long ans;
  std::map<long long, long long > mp;
  while(right-left>1){
    ans = 0;
    mid = (left+right)/2;
    if (get_possible_count(mid)>=M){ // 
      left = mid;
    } else {
      right = mid;
    }
    mp[mid] = get_possible_count(mid);
  }

  long long ans_temp;//= 0;
  if (left == mid){
    for (int i=0; i<N; i++){
      ans_temp += souwa[kumisuu[std::max(left-A[i],(long long)0)]];
      ans_temp += A[i]*kumisuu[std::max(left-A[i],(long long)0)];
    }
    std::cout << ans_temp-abs(M-mp[left])*left << std::endl;
  } else if (right==mid){
    for (int i=0; i<N; i++){
      ans_temp += souwa[kumisuu[std::max(right-A[i],(long long)0)]];
      ans_temp += A[i]*kumisuu[std::max(right-A[i],(long long)0)];
    }
    std::cout << ans_temp + abs(M-mp[right])*left << std::endl;
  }
  return 0;
}