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"
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;

using ll = long long;

int main() {
    int N, abssum = 0, amax = -10001, amin = 10001;
    cin >> N;
    vector<int> A(N);
    rep(i, N) {
        cin >> A[i];
        abssum += abs(A[i]);
        amax = max(amax, A[i]);
        amin = min(amin, A[i]);
    }
    cout << abssum - abs(amax) - abs(amin) + abs(amax - amin) << endl;
    sort(A.begin(), A.end());
    int i = 0, j = N - 1;
    while (i + 1 < j) {
        if (A[i + 1] > 0) {
            cout << A[i] << " " << A[j] << endl;
            A[i] -= A[j];
            j--;
        } else {
            cout << A[j] << " " << A[i] << endl;
            A[j] -= A[i];
            i++;
        }
    }
    cout << A[j] << " " << A[i] << endl;
}
