#include <bits/stdc++.h>
using namespace std;
const long long INFLL = 0x7f7f7f7f7f7f7f7f, MOD = 1e9 + 7;
const int INF = 0x7f7f7f7f;
vector<vector<int>> adj;
vector<int> vis, color;
int dx8[] = {0, 1, 1, 1, 0, -1, -1, -1}, dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
    dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
inline long long mexp(long long x, long long n, long long m = MOD) {
  long long res = 1;
  x %= m;
  while (n) {
    if (n & 1) res = (res * x) % m;
    n >>= 1;
    x = (x * x) % m;
  }
  return res;
}
inline long long gcd(long long a, long long b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
int32_t main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed;
  cerr << fixed;
  cout << setprecision(10);
  cerr << setprecision(3);
  mt19937 genr(chrono::high_resolution_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i];
  sort(v.begin(), v.end());
  long long ans = 0;
  for (int i = 0; i < n / 2; ++i) {
    int temp = (v[i] + v[n - 1 - i]);
    ans += 1LL * temp * temp;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

Lunar New Year is approaching, and Bob is struggling with his homework – a number division problem.

There are n positive integers a_1, a_2, …, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$∑_{j = 1}^{m} s_j^2.$$$

Bob is puzzled by this hard problem. Could you please help him solve it?

Input

The first line contains an even integer n (2 ≤ n ≤ 3 ⋅ 10^5), denoting that there are n integers on Bob's homework paper.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^4), describing the numbers you need to deal with.

Output

A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$∑_{i = j}^{m} s_j^2, where m$$$ is the number of groups.

Examples

Input


4
8 5 2 3


Output


164


Input


6
1 1 1 2 2 2


Output


27

Note

In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.

In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INFLL = 0x7f7f7f7f7f7f7f7f, MOD = 1e9 + 7;
const int INF = 0x7f7f7f7f;
vector<vector<int>> adj;
vector<int> vis, color;
int dx8[] = {0, 1, 1, 1, 0, -1, -1, -1}, dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
    dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
inline long long mexp(long long x, long long n, long long m = MOD) {
  long long res = 1;
  x %= m;
  while (n) {
    if (n & 1) res = (res * x) % m;
    n >>= 1;
    x = (x * x) % m;
  }
  return res;
}
inline long long gcd(long long a, long long b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
int32_t main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed;
  cerr << fixed;
  cout << setprecision(10);
  cerr << setprecision(3);
  mt19937 genr(chrono::high_resolution_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i];
  sort(v.begin(), v.end());
  long long ans = 0;
  for (int i = 0; i < n / 2; ++i) {
    int temp = (v[i] + v[n - 1 - i]);
    ans += 1LL * temp * temp;
  }
  cout << ans << endl;
  return 0;
}

```