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;

void out(ll x, ll y){ cout << x << " " << y << endl;}
int main(){
    int n;
    cin >> n;
    vector<int> a(n);
    ll sm = 0;
    for(int i=0;i<n;i++){
        cin >> a[i];
        sm += abs(a[i]);
    }
    sort(a.begin(),a.end());
    if(a[0] < 0 && a[n-1] > 0){
        // マイナスの物はn-1に, プラスの物は0と演算することで絶対値の総和が解
        cout << sm << endl;
    }else{
        // 全符号が同じなら最小値分差っ引く
        if(a[0] < 0) cout << sm - 2*abs(a[n-1]) << endl;
        else cout << sm - 2*abs(a[0]) << endl;
    }
    for(int i=1;i<n-1;i++){
        if(a[i] < 0){
            out(a[n-1], a[i]);
            a[n-1] -= a[i];
        }else{
            out(a[0], a[i]);
            a[0]  -= a[i];
        }
    }
    out(a[n-1], a[0]);
}