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;
using ll = long long;
 
 
int main()
{
    int n; cin >> n; vector<int> a(n); for(int i = 0; i < n; i++)cin >> a[i]; sort(a.begin(), a.end());
 
    int cur = a.front();
    vector<pair<int,int>> rs;
 
    for(int i = 1; i < n - 1; i++) {
        if(a[i] >= 0) {
            rs.emplace_back(cur, a[i]);
            cur -= a[i];
        }
    }
    rs.emplace_back(a.back(), cur);
    cur = a.back() - cur;
 
    for(int i = 1; i < n - 1; i++) {
        if(a[i] < 0) {
            rs.emplace_back(cur, a[i]);
            cur -= a[i];
        }
    }
    cout << cur << endl;
    for(auto i : rs)cout << i.first << " " << i.second << endl;
}