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 9.2.1) (judge verdict: Accepted)

/*input
9 73
67597 52981 5828 66249 75177 64141 40773 79105 16076
*/
#include<bits/stdc++.h>
#define ull         unsigned long long
#define ll             long long
#define M             1000000007
#define pb             push_back
#define p_q         priority_queue
#define pii         pair<ll,ll>
#define vi          vector<ll>
#define vii         vector<pii>
#define mi          map<ll,ll>
#define mii         map<pii,ll>
#define all(a)      (a).begin(),(a).end()
#define sz(x)       (ll)x.size()    
#define endl        '\n'
#define gcd(a,b)    __gcd((a),(b))
#define lcm(a,b)    ((a)*(b)) / gcd((a),(b))
#define ios            ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define mp             make_pair
#define lb             lower_bound
#define ub            upper_bound
#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define ini(a,n,b)    for(ll int i=0;i<n;i++) a[i]=0;
#define F            first
#define S            second
#define DB(x) cerr << __LINE__ << ": " << #x << " = " << (x) << endl
#define TIME        cerr << "\nTime elapsed: " << setprecision(5) <<1000.0 * clock() / CLOCKS_PER_SEC << "ms\n";
using namespace std;

bool DEBUG = 0;

ll n,m;
vi a(100005);
vi pre(100005);

bool val(ll mx)
{
    ll count = 0;
    rep(i,0,n)
    {
        int it = lower_bound(a.begin() , a.end() , max(mx - a[i] , 0ll)) - a.begin();
        // cout<<mx<<" "<<i<< " " << a[i] << " " <<max(mx - a[i] , 0ll)<<" "<<it<<"\n";
        count += ((n-it));
    }
    return count>=m;  
}

void solve()
{
    //initialize();
    cin>>n>>m;
    a.resize(n);
    rep(i,0,n)
        cin>>a[i];
    sort(all(a));
    pre[n-1] = a[n-1];
    rep(i,n-1,0)pre[i] = pre[i+1]+ a[i];
    ll l = 0 , h = 200005, mid;
    while((h-l) > 1)
    {
        mid = (h+l)/2;
        if (val(mid))
            l = mid;
        else
            h = mid-1;
    }
    ll x;
    if (val(max(h,l)))
        x = max(h,l);
    else
        x = min(h,l);
    ll ans = 0;
    ll count = 0;
    rep(i,0,n)
    {
        auto it = lower_bound(a.begin() , a.end() , max(x + 1 - a[i] , 0ll)) - a.begin();
        count += ((n-it));
        if (it == n)
            continue;
        ans += ((n-it) * a[i]) + (pre[it]);
    }
    ans += (m - count) * x;
    cout<<ans<<"\n";

}


int main(void)
{
    ios
    int t=1;
    //cin>>t;
    while(t--)
    {
        solve();
    }
}