#include<bits/stdc++.h>

int main(){
    using ll = long long;
    constexpr int MAX_A=100001;
    int N;
    ll M;
    std::cin>>N>>M;
    std::vector<ll> A(N);
    for(int i=0;i<N;++i) std::cin>>A[i];
    std::vector<ll> cnt(MAX_A);
    for(int i=0;i<N;++i) cnt[A[i]]++;
    std::vector<ll> S(MAX_A+1);
    for(int i=0;i<MAX_A;++i) S[i+1]=S[i]+cnt[i];
    int l=0,r=2*MAX_A+1;
    while(l+1<r){
        int X=(l+r)>>1;
        ll res=0;
        for(int i=std::max(X-MAX_A,0);i<MAX_A;++i){
            res+=cnt[i]*(S[MAX_A]-S[std::max(X-i,0)]);
        }
        if(res>=M) l=X;
        else r=X;
    }
    std::vector<ll> SS(MAX_A+1);
    for(ll i=0;i<MAX_A;++i) SS[i+1]=SS[i]+i*cnt[i];
    ll ans=0,res=0;
    for(ll i=std::max(l-MAX_A,0);i<MAX_A;++i){
        ans+=cnt[i]*((S[MAX_A]-S[std::max(l-i,0LL)])*i+SS[MAX_A]-SS[std::max(l-i,0LL)]);
        res+=cnt[i]*(S[MAX_A]-S[std::max(l-i,0LL)]);
    }
    ans-=(res-M)*l;
    std::cout<<ans<<std::endl;
}
### 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>

int main(){
    using ll = long long;
    constexpr int MAX_A=100001;
    int N;
    ll M;
    std::cin>>N>>M;
    std::vector<ll> A(N);
    for(int i=0;i<N;++i) std::cin>>A[i];
    std::vector<ll> cnt(MAX_A);
    for(int i=0;i<N;++i) cnt[A[i]]++;
    std::vector<ll> S(MAX_A+1);
    for(int i=0;i<MAX_A;++i) S[i+1]=S[i]+cnt[i];
    int l=0,r=2*MAX_A+1;
    while(l+1<r){
        int X=(l+r)>>1;
        ll res=0;
        for(int i=std::max(X-MAX_A,0);i<MAX_A;++i){
            res+=cnt[i]*(S[MAX_A]-S[std::max(X-i,0)]);
        }
        if(res>=M) l=X;
        else r=X;
    }
    std::vector<ll> SS(MAX_A+1);
    for(ll i=0;i<MAX_A;++i) SS[i+1]=SS[i]+i*cnt[i];
    ll ans=0,res=0;
    for(ll i=std::max(l-MAX_A,0);i<MAX_A;++i){
        ans+=cnt[i]*((S[MAX_A]-S[std::max(l-i,0LL)])*i+SS[MAX_A]-SS[std::max(l-i,0LL)]);
        res+=cnt[i]*(S[MAX_A]-S[std::max(l-i,0LL)]);
    }
    ans-=(res-M)*l;
    std::cout<<ans<<std::endl;
}
```