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: Wrong Answer)

#include<bits/stdc++.h>
//ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<pll,ll> ppll;
typedef pair<ll,pll> plpl;
typedef tuple<ll,ll,ll> tl;
ll mod=1000000007;
ll mod2=998244353;
ll mod3=1000003;
ll mod4=998244853;
ll inf=1000000000000000000;
double pi=2*acos(0);
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,n,m) for(ll i=n;i>=m;i--)
int dh[4]={1,-1,0,0};
int dw[4]={0,0,1,-1};
int ddh[8]={-1,-1,-1,0,0,1,1,1};
int ddw[8]={-1,0,1,-1,1,-1,0,1};
ll lmax(ll a,ll b){
    if(a<b)return b;
    else return a;
}
ll lmin(ll a,ll b){
    if(a<b)return a;
    else return b;
}
ll gcd(ll a,ll b){
    if(a<b)swap(a,b);
    if(b==0)return a;
    if(a%b==0)return b;
    return gcd(b,a%b);
}
ll Pow(ll n,ll k){
    ll ret=1;
    ll now=n;
    while(k>0){
        if(k&1)ret*=now;
        now*=now;
        k/=2;
    }
    return ret;
}
ll beki(ll n,ll k,ll md){
  ll ret=1;
  ll now=n;
  while(k>0){
    if(k%2==1){
      ret*=now;
      ret%=md;
    }
    now*=now;
    now%=md;
    k/=2;
  }
  return ret;
}
ll gyaku(ll n,ll md){
  return beki(n,md-2,md);
}
vector<complex<long double>> fft(vector<complex<long double>> a,int bo){
    int n=a.size();
    int h=0;
    while((1<<h)<n)h++;
    rep(i,0,n){
        int j=0;
        rep(k,0,h)j|=((i>>k)&1)<<(h-1-k);
        if(i<j)swap(a[i],a[j]);
    }
    for(ll i=1;i<n;i*=2){
        rep(j,0,i){
            complex<long double> w=polar(1.0,(-(2*pi)/(2*i))*j*bo);
            for(ll k=0;k<n;k+=i*2){
                complex<long double> s=a[j+k];
                complex<long double> t=a[j+k+i]*w;
                a[j+k]=s+t;
                a[j+k+i]=s-t;
            }
        }
    }
    if(bo==-1)rep(i,0,n)a[i]/=n;
    return a;
}
vector<complex<long double>> fft(vector<long double> a,int bo){
    vector<complex<long double>> b(a.size());
    rep(i,0,a.size())b[i]=complex<long double>(a[i],0);
    return fft(b,bo);
}
vector<long double> convolve(vector<long double> a,vector<long double> b){
    int s=a.size()+b.size()-1;
    int t=1;
    while(t<s)t*=2;
    a.resize(t);
    b.resize(t);
    vector<complex<long double>> A=fft(a,1);
    vector<complex<long double>> B=fft(b,1);
    rep(i,0,t)A[i]*=B[i];
    A=fft(A,-1);
    rep(i,0,s)a[i]=A[i].real();
    return a;
}
int main(){
    ios::sync_with_stdio(false);cin.tie(0);
    int n,m;cin>>n>>m;
    vector<long double> a(100001),b(100001);
    int cn[100001];
    fill(cn,cn+100001,0);
    rep(i,0,n){
        int t;cin>>t;
        cn[t]++;
       
    }
    rep(i,0,100001){
        a[i]=(long double)cn[i];
        b[i]=(long double)cn[i];
    }
    vector<long double> anss=convolve(a,b);
    ll ans[anss.size()];
    rep(i,0,anss.size())ans[i]=(ll)(anss[i]+0.5);
    ll sum=0;
    ll cnt=0;
    //cout<<anss.size()<<endl;
    rrep(i,anss.size()-1,0){
        if(cnt+ans[i]>m){
            sum+=i*(m-cnt);
            break;
        }
        else{
            sum+=i*ans[i];
            cnt+=ans[i];
        }
    }
    cout<<sum<<endl;
}