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: Time Limit Exceeded)

#include "bits/stdc++.h"

#define MOD 1000000007
#define rep(i, n) for(ll i=0; i < (n); i++)
#define ALL(v) v.begin(),v.end()
#define rALL(v) v.rbegin(),v.rend()
#define DUMP(i, v)for(ll i=0;i<v.size();i++)cout<<v[i]<<" "

using namespace std;
typedef long long int ll;
typedef vector<ll> llvec;
typedef vector<double> dvec;
typedef pair<ll, ll> P;
struct Edge{ll from, to, cost;};
struct node{ll cost, to;
  bool friend operator>(node a, node b){
    return a.cost>b.cost;
  }
};

ll mod(ll a, ll mod){
  ll res = a%mod;
  if(res<0)res=res + mod;
  return res;
}

ll modpow(ll a, ll n, ll mod){
  ll res=1;
  while(n>0){
    if(n&1) res=res*a%mod;
    a=a*a%mod;
    n>>=1;
  }
  return res;
}

ll modinv(ll a, ll mod){
  return modpow(a, mod-2, mod);
}

ll gcd(ll a, ll b){
  ll r = a%b;
  if(r==0) return b;
  else return gcd(b, a%b);
}

bool is_prime(ll n){
  ll i = 2;
  if(n==1)return false;
  if(n==2)return true;
  bool res = true;
  while(i*i <n){
    if(n%i==0){
      res = false;
    }
    i = i+1;
  }

  //if(i==1)res = false;
  if(n%i==0)res=false;
  return res;
}
ll N, M;
llvec A;

bool chk(ll m){
  ll cnt = 0;
  rep(i, N){
    ll a = A[N-i-1];
    cnt += (ll)( A.end() - lower_bound(ALL(A), m-a));
    if(cnt>M){break;}
  }
  return (cnt<=M);
}

/**************************************
 ** A main function starts from here  **
 ***************************************/
int main(){
  cin >> N >> M;
  A=llvec(N);
  ll mm = -1;
  rep(i, N){
    cin >> A[i];
    mm = max(A[i], mm);
  }
  sort(ALL(A));
  
  ll l=0,r=3*mm;
  while(r-l>1){
    ll m=(r+l)/2;
    if(chk(m))r=m;
    else l=m;
  }
  
  ll cnt = 0;
  ll cum = 0;
  rep(i, N){
    ll a = A[N-1-i];
    auto iter = lower_bound(ALL(A),r-a);
    ll tmp = (ll)(A.end()-iter);
    cnt += tmp;
    cum += a*tmp;
    while(iter!=A.end()){
      cum += (*iter);
      iter++;
    }
    //if(cnt==M)break;
  }
  cout << cum + l * (M-cnt);
  //cout << cum;
  return 0;
}