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;

static const int INF=1e9;
static const int MOD=1000000007;
typedef long long ll;
typedef pair<int,int> P;

int main(){
    int N; cin >> N;
    ll a[N];
    for(int i=0;i<N;i++) cin >> a[i];
    sort(a,a+N);
    vector<pair<ll,ll>> ans;
    ll tmp=a[0];
    for(int i=1;i<N-1;i++){
        if(a[i]>=0){
            ans.push_back({tmp,a[i]});
            tmp-=a[i];
        }
    }
    ans.push_back({a[N-1],tmp});
    tmp=a[N-1]-tmp;
    for(int i=1;i<N-1;i++){
        if(a[i]<0){
            ans.push_back({tmp,a[i]});
            tmp-=a[i];
        }
    }
    cout << tmp << endl;
    for(int i=0;i<N-1;i++){
        cout << ans[i].first << " " << ans[i].second << endl;
    }

}
