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)

//#define DEB1 //output check function
//#define DEB2 //same output, input from input.txt

#include <iostream>
#include <algorithm>
#include <string>
#include <fstream>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <thread>
#include <mutex>
#include <iomanip>

#define ZERO(a) memset(a, 0, sizeof(a))
#define MINUS(a) memset(a, 0xff, sizeof(a))
#define INF(a) memset(a, 0x3f, sizeof(a))
#define ALL(t) (t).begin(),(t).end()
#define FILL(a,c) fill(a, a+ sizeof(a)/sizeof(*a),c)
#define FILL2D(A,c) fill(A[0], A[0] +sizeof(A)/sizeof(**A),c)
#define FILL3D(A,c) fill(A[0][0], A[0][0] +sizeof(A)/sizeof(***A),c)
#define REP(i,m) for(int i=0;i<(m);++i)
#define REPN(i,m,in) for(int i=(in);i<(m);++i)
#define IN(a, x, b) (a<=x && x<b)
#define PB push_back

using namespace std;

using LL = long long;
using PAIR = pair<int, int>;
using PAIRLL = pair<LL,LL>;
using VI = vector<int>;
using VVI = vector<VI>;


template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}


#ifdef DEB1
#define DPRINT(x) cout<< #x << "=" << (x) << endl; // print at debug mode
#else
#define DPRINT(x) ;
#endif
#ifdef DEB1
#define DCHECK(x) cout<< "debug check" << #x << endl;
#else
#define DCHECK(x) ;
#endif
 


/*
constant values
*/

//const int dx[] = {1,0,-1,0};
//const int dy[] = {0,1,0,-1};
//const int MX = 31001;

//const ll INF= 1LL<<60 ;//INF = 1001001001001; //1e9

// Main program
void solve(){
    vector<int> vec;
    queue<int> pos,neg,x,y;
    int N,a;
    cin >> N;
    REP(i,N){
        cin >> a;
        vec.PB(a);

        //chmin(min,A[i]);
        //chmax(max,A[i]);
    }
    int minid = distance(vec.begin(),min_element(vec.begin(), vec.end()));
    int maxid = distance(vec.begin(),max_element(vec.begin(), vec.end()));

    if(minid==maxid){
        ++maxid;
    }
    DPRINT(minid)
    DPRINT(maxid)

    REP(i,N){
        int tmp = vec[i];
        if(i!=minid && i !=maxid){
            if (vec[i] >= 0){
                pos.push(tmp);
                DCHECK('pos')
            }
            else{
                neg.push(tmp);
                DCHECK('neg')
            }
        }
    }
    int now = vec[minid];
    while(!pos.empty()){
        x.push(now);
        y.push(pos.front());

        now = now - pos.front();
        pos.pop();
        DCHECK('pos pop')
    }
    int max = vec[maxid];
    x.push(max);
    y.push(now);
    
    now = max - now;

    while(!neg.empty()){
        x.push(now);
        y.push(neg.front());

        now = now - neg.front();
        neg.pop();
        DCHECK('neg pop')
    }
    cout << now << endl;
    while(!x.empty()){
        cout << x.front() << " " << y.front() << endl;
        x.pop();
        y.pop();
    }
}



// ----------------------------------------------
int main() {

#ifdef DEB2
    cout << "DEBUG MODE"  << endl;
	ifstream in("input.txt"); //for debug
	cin.rdbuf(in.rdbuf()); //for debug
#endif

	int T=1; 

	for (int i = 0; i < T; ++i) {
		solve();
	}
}