#include <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n,m;
    cin>>n>>m;
    ll a[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    sort(a,a+n);
    ll ok=0,ng=200002,mid=(ok+ng)/2;
    while(mid!=ok){
        ll sum=0;
        for(int i=0;i<n;i++){
            sum+=n-(lower_bound(a,a+n,mid-a[i])-a);
        }
        ((sum>=m)?ok:ng)=mid;
        mid=(ok+ng)/2;
    }
    ll s[n+1];
    s[0]=0;
    for(int i=0;i<n;i++){
        s[i+1]=s[i]+a[i];
    }
    ll ans=0;
    for(int i=0;i<n;i++){
        ll num=(upper_bound(a,a+n,ok-a[i])-a);
        ans+=s[n]-s[num];
        ans+=(n-num)*a[i];
        m-=(n-num);
    }
    cout << ans+(ok*m)<<endl;

    return 0;
}

### Prompt

Generate 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
#include <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n,m;
    cin>>n>>m;
    ll a[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    sort(a,a+n);
    ll ok=0,ng=200002,mid=(ok+ng)/2;
    while(mid!=ok){
        ll sum=0;
        for(int i=0;i<n;i++){
            sum+=n-(lower_bound(a,a+n,mid-a[i])-a);
        }
        ((sum>=m)?ok:ng)=mid;
        mid=(ok+ng)/2;
    }
    ll s[n+1];
    s[0]=0;
    for(int i=0;i<n;i++){
        s[i+1]=s[i]+a[i];
    }
    ll ans=0;
    for(int i=0;i<n;i++){
        ll num=(upper_bound(a,a+n,ok-a[i])-a);
        ans+=s[n]-s[num];
        ans+=(n-num)*a[i];
        m-=(n-num);
    }
    cout << ans+(ok*m)<<endl;

    return 0;
}

```