#include <iostream>
#include <cstdio>
#include <algorithm>
#define N 100005
using namespace std;

typedef long long ll;
ll n, m, l, r = 1e18, md, s, mn = 1e18, a[N];

ll f(ll p) {
    ll i, j, re = 0;
    for (i = 0, j = n - 1; i < n; i++) {
        for (; j >= 0 && a[i] + a[j] >= p; j--);
        re += n - j - 1;
    }
    return re;
}

int main()
{
    int i, j;
    cin >> n >> m;
    for (i = 0; i < n; i++) scanf("%lld", &a[i]);
    sort(a, a + n);
    while (l < r) {
        md = (l + r + 1) / 2;
        if (f(md) >= m) l = md;
        else r = md - 1;
    }
    for (i = 0, j = n - 1; i < n; i++) {
        for (; j >= 0 && a[i] + a[j] >= l; j--);
        if (j < n - 1) mn = min(mn, a[i] + a[j + 1]);
        s += a[i] * (n - j - 1) * 2;
    }
    cout << s - (f(l) - m) * mn << endl;
    return 0;
}
### Prompt

Generate 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 <iostream>
#include <cstdio>
#include <algorithm>
#define N 100005
using namespace std;

typedef long long ll;
ll n, m, l, r = 1e18, md, s, mn = 1e18, a[N];

ll f(ll p) {
    ll i, j, re = 0;
    for (i = 0, j = n - 1; i < n; i++) {
        for (; j >= 0 && a[i] + a[j] >= p; j--);
        re += n - j - 1;
    }
    return re;
}

int main()
{
    int i, j;
    cin >> n >> m;
    for (i = 0; i < n; i++) scanf("%lld", &a[i]);
    sort(a, a + n);
    while (l < r) {
        md = (l + r + 1) / 2;
        if (f(md) >= m) l = md;
        else r = md - 1;
    }
    for (i = 0, j = n - 1; i < n; i++) {
        for (; j >= 0 && a[i] + a[j] >= l; j--);
        if (j < n - 1) mn = min(mn, a[i] + a[j + 1]);
        s += a[i] * (n - j - 1) * 2;
    }
    cout << s - (f(l) - m) * mn << endl;
    return 0;
}
```