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;
typedef pair<ll,ll> ii;
typedef vector<ll> vi;
typedef vector<ii> vii;
#define REP(i,n) for (ll i = 0; i < n; ++i)
#define FOR(i,m,n) for (ll i = m; i < n; ++i)

const ll MAX = 1e5+10;
const ll MOD = 1e9+10;

ll N;
vi A;

int main() {
    cin >> N;
    A = vi(N);
    REP (i, N) cin >> A[i];
    sort(A.begin(), A.end());

    int a = A[0], b = A[N-1];
    string txt = "";
    FOR(i,1,N-1) {
        if (A[i] < 0) {
            txt += to_string(b) + " " + to_string(A[i]) + "\n";
            b -= A[i];
        } else {
            txt += to_string(a) + " " + to_string(A[i]) + "\n";
            a -= A[i];
        }
    }
    txt += to_string(b) + " " + to_string(a) + "\n";
    cout << b-a << endl;
    cout << txt;
}