#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  ll N; cin >> N;
  ll M; cin >> M;
  vector<ll> A(N, 0);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  sort(A.begin(), A.end());

  ll l = 0, r = A[N - 1] * 2 + 1;
  while (l + 1 < r) {
    ll m = (l + r) / 2;
    ll cnt = 0;
    for (int i = 0; i < N; i++) {
      cnt += (A.end() - lower_bound(A.begin(), A.end(), m - A[i]));
    }
    if (cnt >= M) {
      l = m;
    } else {
      r = m;
    }
  }

  vector<ll> SUM(N, 0);
  SUM[N - 1] = A[N - 1];
  for (int i = N - 2; i >= 0; i--) {
    SUM[i] = SUM[i + 1] + A[i];
  }

  ll ans = 0;
  ll mcnt = 0;
  for (int i = 0; i < N; i++) {
    ll temp = (A.end() - upper_bound(A.begin(), A.end(), l - A[i]));
    mcnt += temp;
    if (temp == 0) continue;
    ans += (A[i] * temp + SUM[N - temp]);
  }

  ans += (M - mcnt) * l;

  cout << ans << '\n';

  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 <bits/stdc++.h>
using namespace std;
using ll = long long;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  ll N; cin >> N;
  ll M; cin >> M;
  vector<ll> A(N, 0);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  sort(A.begin(), A.end());

  ll l = 0, r = A[N - 1] * 2 + 1;
  while (l + 1 < r) {
    ll m = (l + r) / 2;
    ll cnt = 0;
    for (int i = 0; i < N; i++) {
      cnt += (A.end() - lower_bound(A.begin(), A.end(), m - A[i]));
    }
    if (cnt >= M) {
      l = m;
    } else {
      r = m;
    }
  }

  vector<ll> SUM(N, 0);
  SUM[N - 1] = A[N - 1];
  for (int i = N - 2; i >= 0; i--) {
    SUM[i] = SUM[i + 1] + A[i];
  }

  ll ans = 0;
  ll mcnt = 0;
  for (int i = 0; i < N; i++) {
    ll temp = (A.end() - upper_bound(A.begin(), A.end(), l - A[i]));
    mcnt += temp;
    if (temp == 0) continue;
    ans += (A[i] * temp + SUM[N - temp]);
  }

  ans += (M - mcnt) * l;

  cout << ans << '\n';

  return 0;
}
```