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: Runtime Error)

/**
* @author khokharnikunj8
**/
#include <bits/stdc++.h>

using namespace std;

//macros
#define ll long long
#define endl "\n"
#define pb push_back

//debug

string to_string(string s) {
    return '"' + s + '"';
}

string to_string(const char *s) {
    return to_string((string) s);
}

string to_string(bool b) {
    return (b ? "true" : "false");
}

template<typename A, typename B>
string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template<typename A>
string to_string(A v) {
    bool first = true;
    string res = "{";
    for (const auto &x : v) {
        if (!first) {
            res += ", ";
        }
        first = false;
        res += to_string(x);
    }
    res += "}";
    return res;
}

void debug_out() { cerr << endl; }


template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
}

#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)

typedef complex<long double> base;

int rev(int num, int lg_n) {
    int res = 0;
    for (int i = 0; i < lg_n; ++i)
        if (num & (1 << i))
            res |= 1 << (lg_n - 1 - i);
    return res;
}

void fft(vector<base> &a, bool invert) {
    int n = (int) a.size();
    int lg_n = 0;
    while ((1 << lg_n) < n) ++lg_n;

    for (int i = 0; i < n; ++i)
        if (i < rev(i, lg_n))
            swap(a[i], a[rev(i, lg_n)]);

    for (int len = 2; len <= n; len <<= 1) {
        long double ang = 2 * M_PI / len * (invert ? -1 : 1);
        base wlen(cos(ang), sin(ang));
        for (int i = 0; i < n; i += len) {
            base w(1);
            for (int j = 0; j < len / 2; ++j) {
                base u = a[i + j], v = a[i + j + len / 2] * w;
                a[i + j] = u + v;
                a[i + j + len / 2] = u - v;
                w *= wlen;
            }
        }
    }
    if (invert)
        for (int i = 0; i < n; ++i)
            a[i] /= n;
}

vector<ll> multiply(const vector<int> &a, const vector<int> &b) {
    vector<base> fa(a.begin(), a.end()), fb(b.begin(), b.end());
    size_t n = 1;
    while (n < max(a.size(), b.size())) n <<= 1;
    n <<= 1;
    fa.resize(n), fb.resize(n);
    vector<ll> res;
    fft(fa, false), fft(fb, false);
    for (size_t i = 0; i < n; ++i)
        fa[i] *= fb[i];
    fft(fa, true);

    res.resize(n);
    for (size_t i = 0; i < n; ++i)
        res[i] = int(fa[i].real() + 0.5);
    return res;
}

int main() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
    int n, m;
    cin >> n >> m;
    vector<int> ar(100001);
    vector<int> br(100001);
    for (int i = 0; i < n; i++) {
        int temp;
        cin >> temp;
        ar[temp]++;
        br[temp]++;
    }
    vector<ll> res = multiply(ar, br);
    ll ans = 0;
    for (int i = 200000; i >= 0 && m > 0; i--) {
        if (m >= res[i]) {
            ans += (i * res[i]);
            m -= res[i];
        } else {
            ans += (m * ((ll) i));
            m = 0;
        }
    }
    cout << ans << endl;
    return 0;
}