#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;

int main(){
    int n;
    ll m;
    cin >> n >> m;
    vector<int> cnt(200001, 0), v(n);
    for(int i = 0; i < n; i++){
        cin >> v[i];
        cnt[v[i]]++;
    }
    vector<int> sum = cnt;
    for(int i = 100000-1; i >= 0; i--)  sum[i] += sum[i+1];
    auto f = [&](int mid)->ll{
        ll ret = 0;
        for(int i = 100000; i >= 0; i--)    ret += (ll)cnt[i]*sum[max(mid-i,0)];
        return ret;
    };
    int l = 0, r = 200001;
    while(r-l > 1){
        int mid = (l+r)/2;
        if(f(mid) < m)  r = mid;
        else            l = mid;
    }
    vector<ll> cum(200001, 0);
    for(int i = 100000; i >= 0; i--)    cum[i] = cum[i+1]+(ll)i*cnt[i];
    ll ans = 0;
    for(int i = 0; i < n; i++){
        ans += (ll)v[i]*sum[max(0,l-v[i])] + cum[max(0,l-v[i])];
    }
    ans -= l*(f(l)-m);
    cout << ans << endl;
    return 0;
}
### Prompt

Create a solution in Cpp for 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<vector>
#include<algorithm>
using namespace std;
typedef long long ll;

int main(){
    int n;
    ll m;
    cin >> n >> m;
    vector<int> cnt(200001, 0), v(n);
    for(int i = 0; i < n; i++){
        cin >> v[i];
        cnt[v[i]]++;
    }
    vector<int> sum = cnt;
    for(int i = 100000-1; i >= 0; i--)  sum[i] += sum[i+1];
    auto f = [&](int mid)->ll{
        ll ret = 0;
        for(int i = 100000; i >= 0; i--)    ret += (ll)cnt[i]*sum[max(mid-i,0)];
        return ret;
    };
    int l = 0, r = 200001;
    while(r-l > 1){
        int mid = (l+r)/2;
        if(f(mid) < m)  r = mid;
        else            l = mid;
    }
    vector<ll> cum(200001, 0);
    for(int i = 100000; i >= 0; i--)    cum[i] = cum[i+1]+(ll)i*cnt[i];
    ll ans = 0;
    for(int i = 0; i < n; i++){
        ans += (ll)v[i]*sum[max(0,l-v[i])] + cum[max(0,l-v[i])];
    }
    ans -= l*(f(l)-m);
    cout << ans << endl;
    return 0;
}
```