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;
ll mod = 1e9+7;
#define rep(i,n) for(int i = 0; i < (n); ++i)

int main(){
    int n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n) cin >> a[i];
    sort(a.begin(), a.end());
    ll max_sum = a[n-1];
    ll min_sum = a[0];
    vector<pair<ll, ll>> ans;
    rep(i, n-2) {
        if ( a[i+1] >= 0 ) {
            ans.emplace_back(min_sum, a[i+1]);
            min_sum -= a[i+1];
        }
        else {
            ans.emplace_back(max_sum, a[i+1]);
            max_sum -= a[i+1];
        }
    }
    ans.emplace_back(max_sum, min_sum);
    cout << max_sum - min_sum << endl;
    rep(i, n-1) printf("%lld %lld\n", ans[i].first, ans[i].second);
    return 0;
}
