#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+7;
int a[maxn];
ll n,m,tot;
ll sum[maxn];
ll res,ans;
int main()
{
   scanf("%lld%lld",&n,&m);

   for(int i=1;i<=n;i++) scanf("%d",&a[i]);

   sort(a+1,a+1+n);
   for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];

   int l=0,r=2e5+1;
   while(l<=r)
   {
       int mid=(l+r)/2;
       tot=0;
       for(int i=1;i<=n;i++)
       {
         int pos=lower_bound(a+1,a+1+n,mid-a[i])-a;
         tot+=n-pos+1;
       }
       if(tot>=m)
       {
           l=mid+1;
           res=mid;
       }
       else
       {
           r=mid-1;
       }
   }
    tot=0;
   //cout<<ans<<endl;
   for(int i=1;i<=n;i++)
   {
        int  pos=lower_bound(a+1,a+n+1,res-a[i])-a;
        ans+=sum[n]-sum[pos-1]+a[i]*(n-pos+1);
        tot+=n-pos+1;
   }
   printf("%lld\n",ans-res*(tot-m));

}

### Prompt

Your task is to create a CPP solution 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
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+7;
int a[maxn];
ll n,m,tot;
ll sum[maxn];
ll res,ans;
int main()
{
   scanf("%lld%lld",&n,&m);

   for(int i=1;i<=n;i++) scanf("%d",&a[i]);

   sort(a+1,a+1+n);
   for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];

   int l=0,r=2e5+1;
   while(l<=r)
   {
       int mid=(l+r)/2;
       tot=0;
       for(int i=1;i<=n;i++)
       {
         int pos=lower_bound(a+1,a+1+n,mid-a[i])-a;
         tot+=n-pos+1;
       }
       if(tot>=m)
       {
           l=mid+1;
           res=mid;
       }
       else
       {
           r=mid-1;
       }
   }
    tot=0;
   //cout<<ans<<endl;
   for(int i=1;i<=n;i++)
   {
        int  pos=lower_bound(a+1,a+n+1,res-a[i])-a;
        ans+=sum[n]-sum[pos-1]+a[i]*(n-pos+1);
        tot+=n-pos+1;
   }
   printf("%lld\n",ans-res*(tot-m));

}

```