#include<iostream>
#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define IO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
int n, a[100009];
ll m, ans, sum[100009];
ll solve(ll x)
{
    ll cnt=0;
    ans=0;
    for(int i=1; i<=n; i++)
    {
        int id=lower_bound(a+1,a+n+1,x-a[i])-a;
        cnt+=n-id+1;
        if(id<=n)
            ans+=(sum[n]-sum[id-1])+1LL*(n-id+1)*a[i];
    }
    return cnt;
}
int main()
{
    IO
    cin>>n>>m;
    for(int i=1; i<=n; i++)
        cin>>a[i];
    sort(a+1,a+n+1);
    for(int i=1; i<=n; i++)
        sum[i]=sum[i-1]+a[i];
    ll low=1, high=a[n]*2+1;
    while(low<high)
    {
        ll mid=(low+high)/2;
        if(solve(mid)>=m)
            low=mid+1;
        else
            high=mid;
    }
    solve(low);
    cout<<ans+(m-solve(low))*(low-1);
}

### Prompt

In CPP, your task is to solve 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<iostream>
#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define IO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
int n, a[100009];
ll m, ans, sum[100009];
ll solve(ll x)
{
    ll cnt=0;
    ans=0;
    for(int i=1; i<=n; i++)
    {
        int id=lower_bound(a+1,a+n+1,x-a[i])-a;
        cnt+=n-id+1;
        if(id<=n)
            ans+=(sum[n]-sum[id-1])+1LL*(n-id+1)*a[i];
    }
    return cnt;
}
int main()
{
    IO
    cin>>n>>m;
    for(int i=1; i<=n; i++)
        cin>>a[i];
    sort(a+1,a+n+1);
    for(int i=1; i<=n; i++)
        sum[i]=sum[i-1]+a[i];
    ll low=1, high=a[n]*2+1;
    while(low<high)
    {
        ll mid=(low+high)/2;
        if(solve(mid)>=m)
            low=mid+1;
        else
            high=mid;
    }
    solve(low);
    cout<<ans+(m-solve(low))*(low-1);
}

```