#include<iostream>
#include<string>
#include<algorithm>
#include<functional>

using namespace std;

typedef long long ll;

ll N, M, A[100000 + 100];
ll csum[100000 + 1000];

//f(x) := x以上となる握手を全部やって、(何回の握手, その時の幸せ度)になるか
pair<ll, ll> f(ll x) {
	ll sum = 0;
	ll happiness = 0;
	for (int i = 0; i < N; i++) {
		int pos = (lower_bound(A, A + N, x - A[i] - 1, greater<ll>()) - A);
		sum += (lower_bound(A, A + N, x - A[i] - 1, greater<ll>()) - A);
		happiness += (csum[pos] + A[i] * pos);
	}
	return make_pair(sum, happiness);
}

int main() {
	cin >> N >> M;
	for (int i = 0; i < N; i++)cin >> A[i];

	sort(A, A + N, greater<ll>());
	csum[0] = 0;
	for (int i = 0; i < N; i++)csum[i + 1] = csum[i] + A[i];

	ll l = 0, u = 100000000000;
	while (u - l > 1) {
		ll mid = (l + u) / 2;
		if (f(mid).first < M)u = mid;
		else l = mid;
	}
	cout << f(l).second - (f(l).first - M) * l << endl;
	return 0;
}
### Prompt

Your task is to create 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<iostream>
#include<string>
#include<algorithm>
#include<functional>

using namespace std;

typedef long long ll;

ll N, M, A[100000 + 100];
ll csum[100000 + 1000];

//f(x) := x以上となる握手を全部やって、(何回の握手, その時の幸せ度)になるか
pair<ll, ll> f(ll x) {
	ll sum = 0;
	ll happiness = 0;
	for (int i = 0; i < N; i++) {
		int pos = (lower_bound(A, A + N, x - A[i] - 1, greater<ll>()) - A);
		sum += (lower_bound(A, A + N, x - A[i] - 1, greater<ll>()) - A);
		happiness += (csum[pos] + A[i] * pos);
	}
	return make_pair(sum, happiness);
}

int main() {
	cin >> N >> M;
	for (int i = 0; i < N; i++)cin >> A[i];

	sort(A, A + N, greater<ll>());
	csum[0] = 0;
	for (int i = 0; i < N; i++)csum[i + 1] = csum[i] + A[i];

	ll l = 0, u = 100000000000;
	while (u - l > 1) {
		ll mid = (l + u) / 2;
		if (f(mid).first < M)u = mid;
		else l = mid;
	}
	cout << f(l).second - (f(l).first - M) * l << endl;
	return 0;
}
```