#include<bits/stdc++.h>
using namespace std;
long long n,m;
vector<long long> box,sum;

bool check (long long mid) {
  long long cnt=0;
  for(int i=0;i<n;i++) {
    cnt+=n-(lower_bound(box.begin(),box.end(),mid-box.at(i))-box.begin());
  }
  if(cnt<m) {
    return false;
  }
  else {
    return true;
  }
}

int main() {
  cin >> n >> m;
  box.resize(n,0);
  sum.resize(n+1,0);
  for(int i=0;i<n;i++) {
    cin >> box.at(i);
  }
  sort(box.begin(),box.end());
  for(int i=0;i<n;i++) {
    sum.at(i+1)=sum.at(i)+box.at(n-1-i);
  }
  long long ok=0,ng=1e12;
  long long mid=(ng+ok)/2;
  while(ng-ok>1) {
    if(check(mid)) {
      ok=mid;
    }
    else {
      ng=mid;
    }
    mid=(ng+ok)/2;
  }
  long long ans=0,cnt=0;
  for(int i=0;i<n;i++) {
    int temp=n-(lower_bound(box.begin(),box.end(),ok-box.at(i))-box.begin());
    cnt+=temp;
    ans+=box.at(i)*temp+sum.at(temp);
  }
  ans-=ok*(cnt-m);
  cout << ans << endl;
}

### 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<bits/stdc++.h>
using namespace std;
long long n,m;
vector<long long> box,sum;

bool check (long long mid) {
  long long cnt=0;
  for(int i=0;i<n;i++) {
    cnt+=n-(lower_bound(box.begin(),box.end(),mid-box.at(i))-box.begin());
  }
  if(cnt<m) {
    return false;
  }
  else {
    return true;
  }
}

int main() {
  cin >> n >> m;
  box.resize(n,0);
  sum.resize(n+1,0);
  for(int i=0;i<n;i++) {
    cin >> box.at(i);
  }
  sort(box.begin(),box.end());
  for(int i=0;i<n;i++) {
    sum.at(i+1)=sum.at(i)+box.at(n-1-i);
  }
  long long ok=0,ng=1e12;
  long long mid=(ng+ok)/2;
  while(ng-ok>1) {
    if(check(mid)) {
      ok=mid;
    }
    else {
      ng=mid;
    }
    mid=(ng+ok)/2;
  }
  long long ans=0,cnt=0;
  for(int i=0;i<n;i++) {
    int temp=n-(lower_bound(box.begin(),box.end(),ok-box.at(i))-box.begin());
    cnt+=temp;
    ans+=box.at(i)*temp+sum.at(temp);
  }
  ans-=ok*(cnt-m);
  cout << ans << endl;
}

```