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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(long long int*)b - *(long long int*)a;} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b){return *(char*)a - *(char*)b;} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b){return strcmp(*(const char **)a, *(const char **)b);} /* aaa,aab.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 100001
#define MOD 1000000007

typedef long long int lli;

 
// 優先度付きキュー Priority Queue

typedef struct
{
  long long int pri;
  long long int left;
  long long int right;
} que;

que pq[MAX];
int pq_size = 0; // 最大要素

void PQadd(long long int pri, long long int left, long long int right)
{
  int last = ++pq_size;
  que new;
  new.pri = pri;
  new.left = left;
  new.right = right;
  pq[pq_size] = new;

  while (last > 1 && pq[last].pri > pq[last / 2].pri)
    {
      que temp;
      temp = pq[last];
      pq[last] = pq[last / 2];
      pq[last / 2] = temp;
      last /= 2;
    }
  return;
}

que PQget(void)
{
  que root = pq[1];
  pq[1] = pq[pq_size--];

  int last = 1, bigger_child;
  while (last * 2 <= pq_size)
    {
      if (last * 2 == pq_size)
	bigger_child = last * 2;
      else
	bigger_child = pq[last * 2].pri > pq[last * 2 + 1].pri ? last * 2 : last * 2 + 1;

      if (pq[last].pri < pq[bigger_child].pri)
	{
	  que temp;
	  temp = pq[last];
	  pq[last] = pq[bigger_child];
	  pq[bigger_child] = temp;
	  last = bigger_child;
	}
      else
	break;
    }
  return root;
}


int main(void) {
  lli n, m;
  scanf("%lld %lld", &n, &m);
  lli a[MAX];
  rep(i, 0, n) scanf("%lld", &a[i]);

  qsort(a, n, sizeof(lli), des);

  
  lli ans = 0;
  lli prev = 0;
  PQadd(a[0] * 2, 0, 0);
  for (lli mi = 0; mi < m; mi++) {
    lli left = pq[1].left;
    lli right = pq[1].right;
    lli max = pq[1].pri;
    PQget();
    ans += max;
    PQadd(a[left + 1] + a[right], left + 1, right);
    if (left != right)
      PQadd(a[left] + a[right + 1], left, right + 1);
    if (right == 0)
      ans += max;
  }
  printf("%lld\n", ans);
  return 0;
}