Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

5 3
10 14 19 34 33

Sample Output 1

202

Let us say that Takahashi performs the following handshakes:

In the first handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 4.

In the second handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 5.

In the third handshake, Takahashi shakes the left hand of Guest 5 and the right hand of Guest 4.

Then, we will have the happiness of (34+34)+(34+33)+(33+34)=202.

We cannot achieve the happiness of 203 or greater, so the answer is 202.

Sample Input 2

9 14
1 3 5 110 24 21 34 5 3

Sample Output 2

1837

Sample Input 3

9 73
67597 52981 5828 66249 75177 64141 40773 79105 16076

Sample Output 3

8128170

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define INF INT_MAX
#define UINF UINT_MAX
#define MOD 1000000007
#define REP(i,n) for(int i=0;i<n;i++)
#define all(x) (x).begin(),(x).end()

typedef long long ll;
typedef long double ld;
typedef std::pair<int,int> pii;

using namespace std;

// Cooley–Tukey FFT algorithm O(N log N)
vector<complex<ld>> fft(vector<complex<ld>> a, bool inverse = false) {
    int n = a.size();
    int h = 0; // h = log_2(n)
    for (int i = 0; 1 << i < n; i++) h++;
    // バタフライ演算用の配置入れ替え
    for (int i = 0; i < n; i++) {
        int j = 0;
        for (int k = 0; k < h; k++) j |= (i >> k & 1) << (h - 1 - k);
        if (i < j) swap(a[i], a[j]);
    }
    // バタフライ演算
    for (int b = 1; b < n; b *= 2) {
        // 第 log_2(b) + 1 段
        // ブロックサイズ = b * 2
        for (int j = 0; j < b; j++) {
            // ブロック内 j 個目
            // 重み w = (1 の原始 2b 乗根の j 乗)
            complex<ld> w =
                polar(1.0, (2 * M_PI) / (2 * b) * j * (inverse ? 1 : -1));
            for (int k = 0; k < n; k += b * 2) {
                // k を先頭とするブロック
                complex<ld> s = a[j + k];         // 前
                complex<ld> t = a[j + k + b] * w; // 後
                a[j + k] = s + t;                     // 前の更新
                a[j + k + b] = s - t;                 // 後の更新
            }
        }
    }
    // 逆変換時にサイズで割る調整
    if (inverse)
        for (int i = 0; i < n; i++) a[i] /= n;
    return a;
}
// Cooley–Tukey FFT algorithm O(N log N)
vector<complex<ld>> fft(vector<ld> a, bool inverse = false) {
    vector<complex<ld>> a_complex(a.size());
    for (int i = 0; i < a.size(); i++) a_complex[i] = complex<ld>(a[i], 0);
    return fft(a_complex, inverse);
}
 
// FFT による畳み込み O(N log N)
vector<ld> convolve(vector<ld> a, vector<ld> b) {
    int s = a.size() + b.size() - 1; // 畳み込み結果のサイズ
    int t = 1; // FFT に使う配列のサイズ（2 の累乗）
    while (t < s) t *= 2;
    a.resize(t); // FFT するためにリサイズ
    b.resize(t); // FFT するためにリサイズ
    vector<complex<ld>> A = fft(a);
    vector<complex<ld>> B = fft(b);
    for (int i = 0; i < t; i++) {
        A[i] *= B[i]; // 畳み込み結果の FFT 結果を得る
    }
    A = fft(A, true); // IFFT で畳み込み結果を得る
    a.resize(s);      // 畳み込み結果を入れるためにリサイズ
    for (int i = 0; i < s; i++) a[i] = A[i].real(); // 実部が答え
    return a;
}

int main() {
    ll max_num = 100100;
    int N,M;
    cin >> N >> M;
    vector<ld> a(max_num+1,0);
    a[0] = 0; // 0 円のメニューは存在しない
    for (int i = 0; i < N; i++) {
        int aa = 0;
        cin >> aa;
        a[aa]+=1.0; 
    }
 
    // 畳み込みで答えを計算
    vector<ld> ret = convolve(a, a);
 
    //ld → int
    vector<int> ret_int(ret.size(),0);
    for(int i = 0;i < ret.size();i++){
        ret_int[i] = (int)(ret[i] + 0.5);
    }

    ll ans = 0;
    ll count = 0;

    for (ll i = ret_int.size() - 1; i > 0 && count < M; i--) {
        if(ret_int[i]){
            // for(int j = 0;j < ret_int[i] && count < M;j++){
            //     ans += i;
            //     count++;
            // }
            if(M >= count + (ll)ret_int[i]){
                ans += (ll)ret_int[i] * i;
                count += (ll)ret_int[i];
            }else{
                ans += (ll)(M - count) * i ;
                break;
            }
        }
    }

    cout << ans << endl;
}