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: Accepted)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
#include <unordered_set>
#include <unordered_map>
#include <iomanip>
//#pragma GCC optimize("Ofast")  //デバッグ時にoptimized-outされて邪魔だった
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef pair<int,int> pii;
typedef vector<pair<int,int> > vpii;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<ll> > vvll;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define irep(it, stl) for(auto it = stl.begin(); it != stl.end(); it++)
#define drep(i,n) for(int i = (n) - 1; i >= 0; --i)
#define fin(ans) cout << (ans) << '\n'
#define mp(p,q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define floatprec(dig) fixed << setprecision(dig)
#define Sort(a) sort(a.begin(), a.end())
#define Rort(a) sort(a.rbegin(), a.rend())
#define MATHPI acos(-1)
#define itn int;
#define invar(typ, var) typ var; cin >> var;
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
template <class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template <class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
struct io{io(){ios::sync_with_stdio(false);cin.tie(0);}};
const int INF = INT_MAX / 2;
const ll LLINF = 1LL<<60;
constexpr ll MOD = 1000000007;
const double EPS = 1e-9;

//増加点数がX以上の組はいくつあるか
ll solve(vi &A, ll X) {
  //まず左を決める
  ll res = 0;
  for(auto itr = A.begin(); itr != A.end(); itr++) {
    ll left = *itr;
    //右
    //A全体を検索する（左右が違っていれば別とみなすので）
    auto right = lower_bound(A.begin(), A.end(), X - left);
    if(right == A.end()) {
      //ない
      continue;
    }
    
    res += A.size() - (right - A.begin());
  }
  
  return res;
}
//67以上になるはず
//res=3

signed main(void) {
  cin.tie(0); ios::sync_with_stdio(false);
  
  ll N, M;
  cin >> N >> M;
  vi A(N);
  rep(i, N) cin >> A[i];
  
  //昇順ソート
  sort(all(A));
  
  //増加点数の最小値で二分探索
  ll lb = -1;
  ll ub = 10000000000LL;
  while(ub - lb > 1) {
    ll md = lb + (ub - lb) / 2LL;
    ll res1 = solve(A, md);
    if(res1 >= M) {
      //M組以上存在する最小の「増加点数の最小値」を取得したい
      lb = md;
    } else {
      ub = md;
    }
  }
  
  //結果
  ll upv = lb;
  
  //upv + 1以上増える組数を取得 + 合計値を取得
  
  ll X = upv + 1;
  //累積和（大きい方，右から）
  vll acc(N+1);
  for(ll i=N-1; i>=0; i--){
    acc[i] = acc[i+1] + A[i];
  }
  
  //まず左を決める
  ll resv = 0;
  ll kumi = 0;
  for(auto itr = A.begin(); itr != A.end(); itr++) {
    ll left = *itr;
    //右
    //?
    auto right = lower_bound(A.begin(), A.end(), X - left);
    if(right == A.end()) {
      //ない
      continue;
    }
    
    ll foundIndex = right - A.begin();
    resv += acc[foundIndex] + left * (A.size() - foundIndex);
    kumi += A.size() - foundIndex;
  }
  
  ll addi = M - kumi;
  resv += addi * upv;
  
  fin(resv);
}