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}

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}

Examples

Input

3
1 -1 2


Output

4
-1 1
2 -2


Input

3
1 1 1


Output

1
1 1
1 0
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;



int main(){

    ll N;
    cin >> N;
    vector<ll>A(N);
    for(ll i = 0; i < N; i++)cin >> A[i];
    A.push_back(-191919);
    sort(A.begin(), A.end());
    A[0] = A[N];
    A[N] = -1;
    ll M = A[0];
    vector<ll>X, Y;
    for(ll i = 1; i < N; i++){
        if(A[i + 1] > 0){
            X.push_back(min(M, A[i]));
            Y.push_back(max(M, A[i]));
            M = -abs(M - A[i]);
        }
        else{
            X.push_back(max(M, A[i]));
            Y.push_back(min(M, A[i]));
            M = abs(M - A[i]);
        }
    }
    cout << M << endl;
    for(ll i = 0; i < X.size(); i++){
        cout << X[i] << " " << Y[i] << endl;
    }
    
    return 0;
}