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 N;
int sum;
vector<int> as;
int main(){
    cin >> N;
    for(int i = 0; i < N; i++){
        int A;
        cin >> A;
        as.push_back(A);
        sum += A;
    }
    sort(as.rbegin(), as.rend());

    sum -= 2* as[N - 1];
    for(int i = N - 2; i > 0; i--){
        if(as[i] < 0) sum -= 2 * as[i];
    }

    cout << sum << endl;

    int idx = 1;
    int Y = as[N - 1];
    for(;idx < N - 1; idx++){
        if(as[idx] < 0) break;
        cout << Y << " " << as[idx] << endl;
        Y -= as[idx];
    }

    int X = as[0];
    for(;idx < N - 1; idx++){
        cout << X << " " << as[idx] << endl;
        X -= as[idx]; 
    }
    cout << X << " " << Y << endl;
    return 0;
}