#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define mp make_pair
#define pb push_back
#define ip pair<ll,ll>
#define ff first
#define ss second
#define MAX 200005
const ll MOD = 1000000007;

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0);

	ll n,m;
	cin>>n>>m;
	ll a[n];
	for(int i=0;i<n;i++)
		cin>>a[i];
	sort(a,a+n);
	ll pre[n+1];
	pre[0] = 0;
	for(int i=1;i<=n;i++)
		pre[i] = pre[i-1]+a[i-1];
	ll lo = 0;
	ll hi = MAX;
	ll ans;
	while(lo<hi){
		ll mid = (lo+hi)/2;
		ll shakes = 0;
		ll j=0;
		ans = 0;
		for(ll i=n-1;i>=0;i--){
			while(j<=i && a[i]+a[j]<mid)
				++j;
			if(j>i)
				break;
			shakes += 2*(i-j)+1;
			ans += 2*(i+1-j)*a[i];
			ans += 2*(pre[i]-pre[j]);
		}

		if(shakes >= m){
			ans -= mid*(shakes-m);
			lo = mid+1;
		}
		else{
			ans += (mid-1)*(m-shakes);
			hi = mid;
		}
	}

	cout<<ans<<endl;
	// lo is smallest value not possible
}
### Prompt

Please provide a CPP coded solution 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;
typedef long long ll;
typedef long double ld;
#define mp make_pair
#define pb push_back
#define ip pair<ll,ll>
#define ff first
#define ss second
#define MAX 200005
const ll MOD = 1000000007;

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0);

	ll n,m;
	cin>>n>>m;
	ll a[n];
	for(int i=0;i<n;i++)
		cin>>a[i];
	sort(a,a+n);
	ll pre[n+1];
	pre[0] = 0;
	for(int i=1;i<=n;i++)
		pre[i] = pre[i-1]+a[i-1];
	ll lo = 0;
	ll hi = MAX;
	ll ans;
	while(lo<hi){
		ll mid = (lo+hi)/2;
		ll shakes = 0;
		ll j=0;
		ans = 0;
		for(ll i=n-1;i>=0;i--){
			while(j<=i && a[i]+a[j]<mid)
				++j;
			if(j>i)
				break;
			shakes += 2*(i-j)+1;
			ans += 2*(i+1-j)*a[i];
			ans += 2*(pre[i]-pre[j]);
		}

		if(shakes >= m){
			ans -= mid*(shakes-m);
			lo = mid+1;
		}
		else{
			ans += (mid-1)*(m-shakes);
			hi = mid;
		}
	}

	cout<<ans<<endl;
	// lo is smallest value not possible
}
```