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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

#define pb push_back
#define ff first
#define ss second
#define all(a) a.begin(), a.end()
#define sz(a) (int)(a.size())
#define clr(a, b) memset(a, b, sizeof(a))
#define REP(i, a, b) for(int i=(int)(a); i<(int)(b); ++i)
#define REPD(i, a, b) for(int i=(int)(a)-1; i>=(int)(b); --i)

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int MX = 1e5 + 5;

int N, a[MX];

int sgn(int aa) {
    if(aa < 0)
        return -1;
    return 1;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> N;
    REP(i, 0, N) {
        cin >> a[i];
    }
    sort(a, a+N);
    
    vector<pair<int, int>> res; 

    if(sgn(a[0]) == sgn(a[N-1])) {
        if(sgn(a[0]) == -1) reverse(a, a+N);
        int cur = a[0] - a[1];
        res.pb({a[0], a[1]});
        a[1] = cur;
        REP(i, 1, N-1) {
            if(i == N-2 && sgn(a[N-1]) == 1) {
                res.pb({a[i+1], a[i]});
            } else {
                res.pb({a[i], a[i+1]});
            }
            cur = cur - a[i+1];
            a[i+1] = cur;
        }
    } else {
        int ind = 0;
        while(sgn(a[ind]) == -1) {
            ++ind;
        }
        --ind;
        int cur = a[ind];
        REP(i, ind, N-1) {
            res.pb({a[i], a[i+1]});
            cur = cur - a[i+1];
            a[i+1] = cur;
        } 
        swap(res.back().ff, res.back().ss);
        cur = res.back().ff - res.back().ss;
        REP(i, 0, ind) {
            res.pb({cur, a[i]});
            cur = cur - a[i];
        }
    }

    assert(sz(res) == N-1);

    cout << res[N-2].ff - res[N-2].ss << "\n";

    for(auto x: res) {
        cout << x.ff << " " << x.ss << "\n";
    }
    return 0;
}