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;
typedef long long ll;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)

int main(){
    ll N;
    cin >> N;
    vector<pair<ll,ll>> a(N);
    rep(i, 0, N){
        cin >> a[i].first;
        a[i].second = i;
    }
    sort(a.begin(), a.end());
    ll neg = a[0].first;
    ll pos = a[N-1].first;
    vector<ll> ans;
    rep(i, 1, N-1){
        if(a[i].first < 0){
            ans.push_back(pos);
            pos -= a[i].first;
        }else{
            ans.push_back(neg);
            neg -= a[i].first;
        }
    }
    cout << pos - neg << endl;
    rep(i, 1, N-1){
        cout << ans[i-1] << " " << a[i].first << endl;
    }
    cout << pos << " " << neg << endl;
    return 0;
}