//https://codeforces.com/contest/1283/problem/B
//BROWNIE TK

#include <bits/stdc++.h>
#include <fstream>
#include <sstream>
#include <complex>
typedef long long int lli;
typedef unsigned long long int ulli;
typedef long double ldb;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

#define pb push_back
#define popb pop_back()
#define pf push_front
#define popf pop_front()

#define si size()
#define be begin()
#define en end()
#define all(v) v.be, v.en

#define le length()
#define mp make_pair
#define mt make_tuple
#define acc(v) accumulate(all(v), 0)
#define F first
#define S second

#define pll pair<lli, lli>
#define pii pair<int, int>
#define pil pair<int, lli>

#define forz(i, n) for (int i = 0; i < n; i++)
#define fore(i, m, n) for (int i = m; i <= n; i++)
#define rforz(i, n) for (int i = n - 1; i >= 0; i--)
#define rfore(i, m, n) for (int i = n; i >= m; i--)

#define deci(n) fixed << setprecision(n)
#define high(n) __builtin_popcount(n)
#define highll(n) __builtin_popcountll(n)
#define parity(n) __builtin_parity(n)
#define ctz(n) __builtin_ctz(n)

#define bset(a, p) ((a) | (1ll << (p)))
#define bchk(a, p) ((a) & (1ll << (p)))
#define bxor(a, p) ((a) ^ (1ll << (p)));
#define brem(a, p) (bchk(a, p) ? (bxor(a, p)) : (a))
/*SOME BITMASK KNOWLEDGE
1)x & (x - 1):sets the last one bit of x to zero
power of two exactly when x & (x − 1) = 0.
2)x & -x:sets all the one bits to zero, except last one bit
3)x | (x - 1):inverts all the bits after the last one bit*/

#define lb lower_bound
#define ub upper_bound
#define er equal_range

#define findnot find_first_not_of

#define maxe *max_element
#define mine *min_element

#define mod 1000000007
#define mod2 998244353
#define gcd __gcd
#define kira ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)

#define endl "\n"
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define p2(a, b) cout << a << " " << b << endl
#define p3(a, b, c) cout << a << " " << b << " " << c << endl
#define p4(a, b, c, d) cout << a << " " << b << " " << c << " " << d << endl

#define oset tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define osetlli tree<lli, null_type, less<lli>, rb_tree_tag, tree_order_statistics_node_update>
//member functions :
//1. order_of_key(k) : number of elements sbtriectly lesser than k
//2. find_by_order(k) : k-th element in the set
#define ofk order_of_key
#define fbo find_by_order
using namespace std;

/*STD funcions*/
lli power(lli x, lli y, lli p)
{
    lli res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
lli modi(lli a, lli m) { return power(a, m - 2, m); }
/*CODE BEGINS*/
const ldb PI = acos(-1);
using cd = complex<ldb>;
void fft(vector<cd> &a, bool invert)
{
    lli n = a.size();
    if (n == 1)
        return;

    vector<cd> a0(n / 2), a1(n / 2);
    for (lli i = 0; 2 * i < n; i++)
    {
        a0[i] = a[2 * i];
        a1[i] = a[2 * i + 1];
    }
    fft(a0, invert);
    fft(a1, invert);

    ldb ang = 2 * PI / n * (invert ? -1 : 1);
    cd w(1), wn(cos(ang), sin(ang));
    for (lli i = 0; 2 * i < n; i++)
    {
        a[i] = a0[i] + w * a1[i];
        a[i + n / 2] = a0[i] - w * a1[i];
        if (invert)
        {
            a[i] /= 2;
            a[i + n / 2] /= 2;
        }
        w *= wn;
    }
}
vector<lli> multiply(vector<lli> const &a)
{
    vector<cd> fa(a.begin(), a.end());
    lli n = 1;
    while (n < 2*a.size())
        n <<= 1;
    fa.resize(n);
    fft(fa, false);
    for (lli i = 0; i < n; i++)
        fa[i] *= fa[i];
    fft(fa, true);

    vector<lli> result(n);
    for (lli i = 0; i < n; i++)
        result[i] = round(fa[i].real());
    return result;
}
const lli N=1<<18;
vector<lli>a(N);
int main()
{
    kira;
    lli n;
    lli m;
    cin >> n >> m;
    lli x;
    forz(i, n)
    {
        cin>>x;
        a[x]++;
    }
    vector<lli>ans=multiply(a);
    lli res=0;
    for(lli i=lli(ans.si)-1;i>=0;i--)
    {
        if(m<=0)break;
        lli temp=min(m,ans[i]);
        m-=temp;
        res+=temp*lli(i);
    }
    cout<<res;
    return 0;
}


Measured memory_bytes: 53500