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: Wrong Answer)

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

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int,int> pii;
typedef pair<double, double> pdd;
typedef pair<ll, ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef vector<pdd> vdd;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define mem(a,b) memset(a, b, sizeof(a) )
#define all(x) (x).begin(),(x).end()
#define INF 1000000000000
#define MOD 1000000007
#define PB push_back
#define MP make_pair
#define F first
#define S second

inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

int main(){
    int n;cin>>n;
    vi a(n),b(n,0);
    rep(i,n)cin>>a[i];
    sort(all(a));
    int e;
    rep(i,n){
        if(a[i]>=0){
            e=i;
          break;
        }
    }
    int cnt=0;
    for(int i=e-1;i>=0;--i){
        b[i]=1;
        cnt++;
        if(cnt==n/2)break;
    }
    if(cnt<n/2){
    for(int i=e;i<n;++i){
        b[i]=1;
        cnt++;
        if(cnt==n/2)break;
    }}
    int sum=0;
    rep(i,n){
        if(b[i]==0)sum+=a[i];
        else sum+=-a[i];
    }
    cout<<sum<<endl;
        vi c,d;
    rep(i,n){
        if(b[i]==0)c.PB(a[i]);
        else d.PB(a[i]);
    }
    int t=c[0];
    for(int i=0;i<d.size();++i){
        cout<<d[i]<<" "<<t<<endl;
        t=d[i]-t;
    }
    for(int i=1;i<c.size();++i){
        cout<<c[i]<<" "<<t<<endl;
        t=t-c[i];
    }
}