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"

using namespace std;

using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using vvvll = vector<vector<vector<ll>>>;
using vvvvll = vector<vector<vector<vector<ll>>>>;
using pl4 = pair<ll,ll>;
using str = string;
using vpl4 = vector<pair<ll,ll>>;

#define sz size()
#define be begin()
#define en end()
#define fi first
#define se second
#define pb push_back
#define mp make_pair

#define llin(x) ll (x);cin >>(x);
#define stin(x) str (x);cin >>(x);
#define vllin(x,n) vll (x)(n);FOR(i,0,n-1){cin >>(x)[i];}
#define vllin2(a,b,n) vll (a)(n);vll (b)(n);FOR(i,0,n-1){cin >>(a)[i]>>(b)[i];}
#define vllin3(a,b,c,n) vll (a)(n);vll (b)(n);vll (c)(n);FOR(i,0,n-1){cin >>(a)[i]>>(b)[i]>>(c)[i];}
#define vlling(x,n) (x).assign(n,0);FOR(i,0,n-1){cin >>(x)[i];}
#define vlling2(a,b,n) (a).assign(n,0);(b).assign(n,0);FOR(i,0,n-1){cin >>(a)[i]>>(b)[i];}
#define vlling3(a,b,c,n) (a).assign(n,0);(b).assign(n,0);(c).assign(n,0);FOR(i,0,n-1){cin >>(a)[i]>>(b)[i]>>(c)[i];}
#define vpl4in(x,n) vpl4 (x)((n),mp(0,0));FOR(i,0,n-1){cin >>x[i].fi>>x[i].se;}
#define FOR(i,a,b) for(ll i = a ; i <= b ; i++)
#define rFOR(i,b,a) for(ll i = a; i >= b ; i--)
#define SORT(x) sort(x.be, x.en)
#define rSORT(x) sort(x.rbegin(), x.rend())
#define say(x) cout<<(x);
#define sal(x) cout<<(x)<<endl;
#define says(x) cout<<(x)<<(' ');
#define sas cout<<(' ');
#define sayR(x) cout<<fixed<<setprecision(10)<<(x);
#define salR(x) cout<<fixed<<setprecision(10)<<(x)<<endl;
#define yn(a) cout <<((a)?"yes":"no")<<endl;
#define Yn(a) cout <<((a)?"Yes":"No")<<endl;
#define YN(a) cout <<((a)?"YES":"NO")<<endl;
#define Imp(a) cout <<((a)?"Possible":"Impossible")<<endl;
#define IMP(a) cout <<((a)?"POSSIBLE":"IMPOSSIBLE")<<endl;
#define pow(a,b) ll(pow(a,b))

ll MOD=1000000007;

signed main(){
        llin(n);
        vllin(a,n);
        SORT(a);
        ll c=0;
        FOR(i,0,n-1){
                if(a[i]>0){
                        c++;
                }
        }
        ll m=0;
        ll k=0;
        if(c>n/2+n%2){
                FOR(i,0,n-1){
                        if(i<n/2){
                                m-=a[i];
                        }else{
                                m+=a[i];
                        }
                }
                sal(m)
                if(n%2==1){
                        FOR(i,0,n/2-1){
                                k=a[n-1-i]-k;
                                says(a[i])
                                sal(k)
                                k=a[i]-k;
                                says(a[n-2-i])
                                sal(k)
                        }
                }else{
                        FOR(i,0,n/2-2){
                                k=a[i]-k;
                                says(a[n-1-i])
                                sal(k)
                                k=a[n-1-i]-k;
                                says(a[i+1])
                                sal(k)
                        }
                        k=a[n/2-1]-k;
                        says(a[n/2])
                        sal(k)
                }
        }else{
                FOR(i,0,n-1){
                        m+=abs(a[i]);
                }
                sal(m)
                rSORT(a);
                FOR(i,0,c-2){
                        k=a[i]-k;
                        says(a[c+i])
                        sal(k)
                        k=a[c+i]-k;
                        says(a[i+1])
                        sal(k)
                }
                k=a[c-1]-k;
                FOR(i,c*2-1,n-1){
                        says(k)
                        sal(a[i])
                        k=k-a[i];
                }
        }
}