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;

const int NS = (int)1e5 + 4;
int N;
int arr[NS];
vector < pair < int, int > > ans;

int main(){
    scanf("%d", &N);
    for(int i = 1; i <= N; ++i){
        scanf("%d", arr + i);
    }
    sort(arr + 1, arr + N + 1);
    for(int i = 2; i < N; ++i){
        if(arr[i] < 0){
            ans.push_back(make_pair(arr[N], arr[i]));
            arr[N] = arr[N] - arr[i];
        }
        else{
            ans.push_back(make_pair(arr[1], arr[i]));
            arr[1] = arr[1] - arr[i];
        }
    }
    printf("%d\n", arr[N] - arr[1]);
    for(auto&i:ans){
        printf("%d %d\n", i.first, i.second);
    }
    printf("%d %d\n", arr[N], arr[1]);
    return 0;
}
