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

using i64 = std::int_fast64_t;

#define REP(i, stop) FOR(i, 0, stop)
#define FOR(i, start, stop) for (int i = start, i##_len = stop; i < i##_len; ++i)
#define RREP(i, n) for (int i = n; i-- > 0;)
#define ALL(v) (v).begin(), (v).end()

signed main() {
  i64 N, M;
  scanf("%lld %lld", &N, &M);
  vector<i64> A(N);
  REP(i, N) scanf("%lld", &A[i]);
  sort(ALL(A));
  vector<i64> S(N + 1);
  REP(i, N) S[i + 1] = S[i] + A[i];
  i64 h_ng = 0;
  i64 h_ok = A[N - 1] * 2 + 1;
  i64 h = h_ok;
  i64 ans = 0;
  while (h_ok > h_ng + 1) {
    h = h_ng + h_ok >> 1;
    i64 s = 0;
    i64 q = 0;
    i64 gt = 0;
    RREP(i, N) {
      while (gt < N && !(A[i] + A[gt] > h)) gt++;
      q += N - gt;
      s += S[N] - S[gt] + A[i] * (N - gt);
    }
    if (q <= M) {
      h_ok = h;
      ans = s + (M - q) * h;
    } else {
      h_ng = h;
    }
  }
  printf("%lld\n", ans);
}

### Prompt

Develop a solution in CPP to the problem described below:

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 i64 = std::int_fast64_t;

#define REP(i, stop) FOR(i, 0, stop)
#define FOR(i, start, stop) for (int i = start, i##_len = stop; i < i##_len; ++i)
#define RREP(i, n) for (int i = n; i-- > 0;)
#define ALL(v) (v).begin(), (v).end()

signed main() {
  i64 N, M;
  scanf("%lld %lld", &N, &M);
  vector<i64> A(N);
  REP(i, N) scanf("%lld", &A[i]);
  sort(ALL(A));
  vector<i64> S(N + 1);
  REP(i, N) S[i + 1] = S[i] + A[i];
  i64 h_ng = 0;
  i64 h_ok = A[N - 1] * 2 + 1;
  i64 h = h_ok;
  i64 ans = 0;
  while (h_ok > h_ng + 1) {
    h = h_ng + h_ok >> 1;
    i64 s = 0;
    i64 q = 0;
    i64 gt = 0;
    RREP(i, N) {
      while (gt < N && !(A[i] + A[gt] > h)) gt++;
      q += N - gt;
      s += S[N] - S[gt] + A[i] * (N - gt);
    }
    if (q <= M) {
      h_ok = h;
      ans = s + (M - q) * h;
    } else {
      h_ng = h;
    }
  }
  printf("%lld\n", ans);
}

```