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>
#define All(v) v.begin(), v.end()
using namespace std;
typedef long long lint;
typedef pair<int, int> pii;
typedef pair<lint, lint> pll;

const int MAX_N = (int)1e5 + 5;
int A[MAX_N], sign[MAX_N];

int main(){
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int N, p=0, m=0, z=0;
    cin>>N;
    for(int i=0;i<N;i++){
        cin>>A[i];
        if(A[i]>0) p++, sign[i] = 1;
        else if(A[i]<0) m++, sign[i] = -1;
        else z++;
    }
    if(m==0){
        int idx = distance(A, min_element(A, A+N));
        sign[idx] = -1;
    }
    else if(p==0){
        int idx = distance(A, max_element(A, A+N));
        sign[idx] = 1;
    }

    int cnt = 0, ans = 0, cur = 0;
    queue<int> P, M;
    for(int i=0;i<N;i++){
        if(sign[i]==1) ans += A[i], P.push(A[i]);
        else ans -= A[i], M.push(A[i]);
    }
    cur = ans;
    vector<int> X, Y;
    while(P.size()>1){
        int x = P.front(); P.pop();
        if(cnt==0){
            cur = x - cur;
            X.push_back(x); Y.push_back(cur);
            cnt ^= 1;
        }
        else{
            cur += x;
            X.push_back(cur); Y.push_back(x);
        }
    }

    while(M.size()>1) {
        int x = M.front(); M.pop();
        if (cnt==1) {
            cur = x - cur;
            X.push_back(x); Y.push_back(cur);
            cnt ^= 1;
        }
        else {
            cur += x;
            X.push_back(cur); Y.push_back(x);
        }
    }

    int xx = P.front(), yy = M.front();
    if(cnt==0){
        X.push_back(xx), Y.push_back(yy);
    }
    else{
        Y.push_back(xx), X.push_back(yy);
    }

    printf("%d\n", ans);
    reverse(All(X)), reverse(All(Y));
    for(int i=0;i<N-1;i++) printf("%d %d\n", X[i], Y[i]);
    return 0;
}