Problem:
Score : 500 points

Problem Statement

There are N integers, A_1, A_2, ..., A_N, written on a blackboard.

We will repeat the following operation N-1 times so that we have only one integer on the blackboard.

Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y.

Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.

Constraints

2 \leq N \leq 10^5

-10^4 \leq A_i \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible value M of the final integer on the blackboard, and a sequence of operations x_i, y_i that maximizes the final integer, in the format below.

Here x_i and y_i represent the integers x and y chosen in the i-th operation, respectively.

If there are multiple sequences of operations that maximize the final integer, any of them will be accepted.

M
x_1 y_1
:
x_{N-1} y_{N-1}

Sample Input 1

3
1 -1 2

Sample Output 1

4
-1 1
2 -2

If we choose x = -1 and y = 1 in the first operation, the set of integers written on the blackboard becomes (-2, 2).

Then, if we choose x = 2 and y = -2 in the second operation, the set of integers written on the blackboard becomes (4).

In this case, we have 4 as the final integer. We cannot end with a greater integer, so the answer is 4.

Sample Input 2

3
1 1 1

Sample Output 2

1
1 1
1 0

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

#include <bits/stdc++.h>

// {{{
#define PI pair<int, int>
#define mk make_pair
#define reg register
#define ll long long
#define rep(i, a, b) for(reg int i = a; i <= b; ++i)
#define per(i, a, b) for(reg int i = a; i >= b; --i)
// }}}

using namespace std;

inline int min(reg int a, reg int b) { return a < b ? a : b; }
inline int max(reg int a, reg int b) { return a > b ? a : b; }
inline int read() {
  reg int s = 0; reg char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
  return s;
}
#define N 100005

int n;
int a[N];
struct Node {
  int val, id;
} b[N];

bool cmp(Node x, Node y) {
  return x.id < y.id;
}

int main() {
  n = read();
  rep (i, 1, n) scanf ("%d", &a[i]);
  sort(a + 1, a + n + 1);
  int tmp = n / 2;
  ll ans = 0;
  rep(i, 1, n) {
    if (i <= tmp) ans -= a[i];
    else ans += a[i];
    b[i].val = a[i];
  }
  cout << ans << endl;
  ll now = 0;
  if (n & 1) {
    rep (i, 1, tmp) {
      b[i].id = 2 * i;
    }
    rep (i, tmp + 1, n) {
      b[i].id = 2 * (i - tmp) - 1;
    }
  }
  else {
    rep (i, 1, tmp) {
      b[i].id = (2 * i - 1);
    }
    rep (i, tmp + 1, n) {
      b[i].id = 2 * (i - tmp);
    }
  }
  sort(b + 1, b + n + 1, cmp);
  now = b[1].val;
  rep (i, 2, n) {
    printf("%d %lld\n", b[i].val, now);
    now = b[i].val - now;
  }
  return 0; 
}



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