#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i=0;i<n;++i)
#define rep1(i,n) for(ll i=1;i<=n;++i)
int main(){
  ll n,m;cin>>n>>m;
  vector<ll> a(n);
  rep(i,n)cin>>a[i];
  sort(a.begin(),a.end());
  ll sum=0;
  ll left=0;ll right=a[n-1]*2+1;
  ll ans=0;
  while(right-left>1){
    ll mid=(left+right)/2;
    sum=0;
    rep(i,n){
      ll buf=mid-a[i];
      auto it=lower_bound(a.begin(),a.end(),buf);
      ll k=it-a.begin();
      sum+=n-k;
    }
    if(sum<=m)right=mid;
    else if(sum>m)left=mid;
  }
//  cout<<right<<endl;
  vector<ll> b(n+1,0);b[n-1]=a[n-1];
  rep1(i,n-1){
    b[n-1-i]=b[n-i]+a[n-1-i];
  }
    rep(i,n){
    ll buf=right-a[i];
    auto it=lower_bound(a.begin(),a.end(),buf);
    ll k=it-a.begin();
    ans+=(n-k)*a[i]+b[k];
    m-=n-k;
  }
  ans+=m*left;
  if(n*n==m){
    ans=(n+1)*b[0];
  }
  std::cout<<ans<<endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

Examples

Input

5 3
10 14 19 34 33


Output

202


Input

9 14
1 3 5 110 24 21 34 5 3


Output

1837


Input

9 73
67597 52981 5828 66249 75177 64141 40773 79105 16076


Output

8128170

### Response

```cpp
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i=0;i<n;++i)
#define rep1(i,n) for(ll i=1;i<=n;++i)
int main(){
  ll n,m;cin>>n>>m;
  vector<ll> a(n);
  rep(i,n)cin>>a[i];
  sort(a.begin(),a.end());
  ll sum=0;
  ll left=0;ll right=a[n-1]*2+1;
  ll ans=0;
  while(right-left>1){
    ll mid=(left+right)/2;
    sum=0;
    rep(i,n){
      ll buf=mid-a[i];
      auto it=lower_bound(a.begin(),a.end(),buf);
      ll k=it-a.begin();
      sum+=n-k;
    }
    if(sum<=m)right=mid;
    else if(sum>m)left=mid;
  }
//  cout<<right<<endl;
  vector<ll> b(n+1,0);b[n-1]=a[n-1];
  rep1(i,n-1){
    b[n-1-i]=b[n-i]+a[n-1-i];
  }
    rep(i,n){
    ll buf=right-a[i];
    auto it=lower_bound(a.begin(),a.end(),buf);
    ll k=it-a.begin();
    ans+=(n-k)*a[i]+b[k];
    m-=n-k;
  }
  ans+=m*left;
  if(n*n==m){
    ans=(n+1)*b[0];
  }
  std::cout<<ans<<endl;
  return 0;
}

```