#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
typedef pair<ll,ll> l_l;
const int INF=1001001000;
const int mINF=-1001001000;
const ll LINF=10100100100100100;
int n;
ll m;
l_l calc(int mid,vector<int> a,vector<ll> sum){
    ll total=0,num=0;
    for(int i=0; i<n; i++){
        int j=lower_bound(a.begin(),a.end(),mid-a[i])-a.begin();
        num+=n-j;
        total+=sum[n]-sum[j];
        total+=a[i]*ll(n-j);
    }
    return l_l(total,num);
}
int main(){
    cin >> n >> m;
    vector<int> a(n);
    for(int i=0; i<n; i++) cin >> a[i];
    vector<ll> sum(n+1);
    sort(a.begin(),a.end());
    for(int i=0; i<n; i++) sum[i+1]=sum[i]+a[i];
    int l=0,r=200005;
    while(r-l>1){
        int mid=(r+l)/2;
        if(calc(mid,a,sum).second>=m) l=mid;
        else r=mid;
    }
    auto p=calc(l,a,sum);
    ll ans=p.first;
    ans-=(p.second-m)*l;
    cout << ans << endl;
    return 0;
}
### Prompt

Please formulate 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>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
typedef pair<ll,ll> l_l;
const int INF=1001001000;
const int mINF=-1001001000;
const ll LINF=10100100100100100;
int n;
ll m;
l_l calc(int mid,vector<int> a,vector<ll> sum){
    ll total=0,num=0;
    for(int i=0; i<n; i++){
        int j=lower_bound(a.begin(),a.end(),mid-a[i])-a.begin();
        num+=n-j;
        total+=sum[n]-sum[j];
        total+=a[i]*ll(n-j);
    }
    return l_l(total,num);
}
int main(){
    cin >> n >> m;
    vector<int> a(n);
    for(int i=0; i<n; i++) cin >> a[i];
    vector<ll> sum(n+1);
    sort(a.begin(),a.end());
    for(int i=0; i<n; i++) sum[i+1]=sum[i]+a[i];
    int l=0,r=200005;
    while(r-l>1){
        int mid=(r+l)/2;
        if(calc(mid,a,sum).second>=m) l=mid;
        else r=mid;
    }
    auto p=calc(l,a,sum);
    ll ans=p.first;
    ans-=(p.second-m)*l;
    cout << ans << endl;
    return 0;
}
```