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;
#define ll long long
const int inf = 0x3f3f3f3f;

ll ans;
int n,a[100005];
int main(){
    cin >> n;
    for(int i=1;i<=n;i++) cin >> a[i];
    sort(a+1,a+n+1);
    for(int i=2;i<n;i++) ans+=abs(a[i]);
    ans-=a[1];ans+=a[n];
    cout << ans << endl;
    int i;
    for(i=2;i<n && a[i]<=0;i++);
    int index=i;
    for(i;i<n;i++){
        cout << a[1] << " " << a[i] << endl;
        a[1]-=a[i];
    }
    for(i=1;i<index;i++){
        cout << a[n] << " " << a[i] << endl;
        a[n]-=a[i];
    }
    return 0;
}
