#include<bits/stdc++.h>
#define int long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
int a[100005],s[100005],n,m; 
int check(int mid){
	int sum=0;
	FOR(i,1,n){
		sum=sum+n-(lower_bound(a+1,a+n+1,mid-a[i])-a)+1;
	}
	return sum;
}
signed main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	cin>>n>>m;
	FOR(i,1,n)cin>>a[i];
	sort(a+1,a+n+1);
	for(int i=n;i>=1;i--)s[i]=s[i+1]+a[i];
	int l=0,r=1000000000,mid,ans=-1;
	while(r>=l){
		mid=(l+r)>>1;
		if(check(mid)>=m){
			ans=mid;l=mid+1;
		}else r=mid-1;
	} 
	int sum=0;
	FOR(i,1,n){
		int t=lower_bound(a+1,a+n+1,ans-a[i])-a;
		sum=sum+a[i]*(n-t+1)+s[t];
	}
	cout<<sum-(check(ans)-m)*ans;
	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<bits/stdc++.h>
#define int long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
int a[100005],s[100005],n,m; 
int check(int mid){
	int sum=0;
	FOR(i,1,n){
		sum=sum+n-(lower_bound(a+1,a+n+1,mid-a[i])-a)+1;
	}
	return sum;
}
signed main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	cin>>n>>m;
	FOR(i,1,n)cin>>a[i];
	sort(a+1,a+n+1);
	for(int i=n;i>=1;i--)s[i]=s[i+1]+a[i];
	int l=0,r=1000000000,mid,ans=-1;
	while(r>=l){
		mid=(l+r)>>1;
		if(check(mid)>=m){
			ans=mid;l=mid+1;
		}else r=mid-1;
	} 
	int sum=0;
	FOR(i,1,n){
		int t=lower_bound(a+1,a+n+1,ans-a[i])-a;
		sum=sum+a[i]*(n-t+1)+s[t];
	}
	cout<<sum-(check(ans)-m)*ans;
	return 0;
}



```