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 <algorithm>
#include <complex>
#include <utility>
#include <vector>
#include <string>
#include <queue>
#include <tuple>
#include <cmath>
#include <bitset>
#include <cctype>
#include <set>
#include <map>
#include <unordered_map>
#include <numeric>
#include <functional>
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(ll i=ll(a);i<ll(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define PRINT(V) cout << V << "\n"
#define SORT(V) sort((V).begin(),(V).end())
#define RSORT(V) sort((V).rbegin(), (V).rend())
using namespace std;
using ll = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline void Yes(bool condition){ if(condition) PRINT("Yes"); else PRINT("No"); }
template<class itr> void cins(itr first,itr last){
    for (auto i = first;i != last;i++){
        cin >> (*i);
    }
}
template<class itr> void array_output(itr start,itr goal){
    string ans = "",k = " ";
    for (auto i = start;i != goal;i++) ans += to_string(*i)+k;
    if (!ans.empty()) ans.pop_back();
    PRINT(ans);
}
ll gcd(ll a, ll b) {
    return a ? gcd(b%a,a) : b;
}
const ll INF = 1e16;
const ll MOD = 1000000007;
typedef pair<ll,ll> P;
const ll MAX = 200010;
constexpr ll nx[8] = {1,0,-1,0,-1,-1,1,1};
constexpr ll ny[8] = {0,1,0,-1,-1,1,-1,1};
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,m;
    cin >> n >> m;
    vector<ll> a(n),sum(n+1);
    cins(all(a));
    SORT(a);
    rep(i,n) sum[i+1] = a[i];
    rep(i,n) sum[i+1] += sum[i];
    ll l = -1,r = 400000;
    while(r-l > 1){
        ll mid = (r+l)/2;
        ll cnt = 0,same = 0;
        rep(i,n){
            if (a[n-1]+a[i] < mid) continue;
            cnt += n-(upper_bound(all(a),mid-a[i])-a.begin());
            same += upper_bound(all(a),mid-a[i])-lower_bound(all(a),mid-a[i]);
        }
        if (cnt <= m && same+cnt >= m){
            l = mid;
            break;
        }
        cnt += same;
        if (cnt > m) l = mid;
        else r = mid;
    }
    ll ans = 0;
    ll cnt = 0;
    rep(i,n){
        if (a[n-1]+a[i] < l) continue;
        //PRINT(l-a[i]);
        ll index = upper_bound(all(a),l-a[i])-a.begin();
        cnt += n-(upper_bound(all(a),l-a[i])-a.begin());
        ans += sum[n]-sum[index]+a[i]*(n-index);
    }
    ans += l*(m-cnt);
    PRINT(ans);
}