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;

int main(){
    long long int N,ans=0,tmpf,tmpb;
    cin >> N;
    vector<long long int> A(N);
    vector<pair<long long int,long long int>> P;

    for(int i=0;i<N;i++){
        cin >> A[i];
    }

    sort(A.begin(),A.end());

    tmpf = A[0];
    tmpb = A.back();

    for(int i=1;i<N-1;i++){
        if(0<=A[i]){
            P.push_back({tmpf,A[i]});
            tmpf -= A[i];
        }else{
            P.push_back({tmpb,A[i]});
            tmpb -= A[i];
        }
    }

    P.push_back({tmpb,tmpf});

    cout << tmpb - tmpf << endl;

    for(int i=0;i<N-1;i++){
        cout << P[i].first << " " << P[i].second << endl;
    }
}