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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n - 1; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i <= n; i++)
#define FORR(i, m, n) for(int i = m; i >= n; i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
 
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<ll, ll>;

int main(){
	cin.tie(0);
    ios::sync_with_stdio(false);

    ll n;
    cin >> n;
    vll a(n);
    ll neg = 0;
    REP(i, n){
        cin >> a[i];
        if(a[i] < 0) neg++;
    }
    if(neg == 0) neg = 1;
    if(neg == n) neg = n - 1;
    VSORT(a);
    vll x(n - 1), y(n - 1);
    REP(i, n - neg - 1){
        x[i] = a[0], y[i] = a[neg + i];
        a[0] -= a[neg + i];
    }
    REP(i, neg){
        x[i + n - neg - 1] = a[n - 1], y[i + n - neg - 1] = a[i];
        a[n - 1] -= a[i];
    }
    cout << a[n - 1] << endl;
    REP(i, n - 1) cout << x[i] << " " << y[i] << endl;
    return 0;
}