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 i64 = int64_t;

int main(){
    int n;
    cin >> n;
    vector<int> v;
    int neg = 0;
    for(int i=0;i<n;++i){
        int a;
        cin >> a;
        v.push_back(a);
        if(a < 0)neg++;
    }
    sort(v.begin(), v.end());
    if(neg == 0) neg++;
    else if(neg == n)neg--;
    vector<pair<int, int>> ans;
    for(int i=neg;i<n-1;++i){
        ans.emplace_back(v[0], v[i]);
        v[0] -= v[i];
    }
    for(int i=0;i<neg;++i){
        ans.emplace_back(v.back(), v[i]);
        v.back() -= v[i];
    }
    cout << v.back() << endl;
    for(auto e: ans){
        cout << e.first << " " << e.second << endl;
    }

    return 0;
}