#include<bits/stdc++.h>
using namespace std;

#define INF (1LL<<60)

long long cnt[200001];
long long sum[200001];
long long A[100000];

long long n,m;

long long Check(long long num){
	long long ret=0;

	for(int i=0;i<n;i++){
		ret+=cnt[ max(num-A[i],0LL) ];
	}
	return ret;
}

int main(){
	
	cin>>n>>m;

	for(int i=0;i<n;i++){
		cin>>A[i];
		cnt[ A[i] ]++;
		sum[ A[i] ]+=A[i];
	}

	// 左手を先に決めた時の、右手握手できる人数を求める用
	for(int i=100000;i>=0;i--){
		cnt[i]+=cnt[i+1];
		sum[i]+=sum[i+1];
	}

	long long OK=0, NG=200001;
	while(NG-OK>1){
		long long mid=(OK+NG)/2, tmp=0;

		if(Check(mid)>=m)OK=mid;
		else NG=mid;
	}
	//cout<<OK<<endl;

	long long ans=0;
	for(int i=0;i<n;i++){
		ans+=sum[ max(OK-A[i],0LL) ] + A[i]*cnt[ max(OK-A[i],0LL) ];
	}
	//cout<<ans<<endl;

	long long tmp=Check(OK);
	//cout<<tmp<<endl;

	ans-=(tmp-m)*OK;
	cout<<ans<<endl;


	return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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;

#define INF (1LL<<60)

long long cnt[200001];
long long sum[200001];
long long A[100000];

long long n,m;

long long Check(long long num){
	long long ret=0;

	for(int i=0;i<n;i++){
		ret+=cnt[ max(num-A[i],0LL) ];
	}
	return ret;
}

int main(){
	
	cin>>n>>m;

	for(int i=0;i<n;i++){
		cin>>A[i];
		cnt[ A[i] ]++;
		sum[ A[i] ]+=A[i];
	}

	// 左手を先に決めた時の、右手握手できる人数を求める用
	for(int i=100000;i>=0;i--){
		cnt[i]+=cnt[i+1];
		sum[i]+=sum[i+1];
	}

	long long OK=0, NG=200001;
	while(NG-OK>1){
		long long mid=(OK+NG)/2, tmp=0;

		if(Check(mid)>=m)OK=mid;
		else NG=mid;
	}
	//cout<<OK<<endl;

	long long ans=0;
	for(int i=0;i<n;i++){
		ans+=sum[ max(OK-A[i],0LL) ] + A[i]*cnt[ max(OK-A[i],0LL) ];
	}
	//cout<<ans<<endl;

	long long tmp=Check(OK);
	//cout<<tmp<<endl;

	ans-=(tmp-m)*OK;
	cout<<ans<<endl;


	return 0;
}

```