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 P = pair<int, int>;

int main() {
    int N; cin >> N;
    vector<int> A(N);
    for(auto &e : A) cin >> e;
    sort(A.begin(), A.end());
    vector<P> ans;
    for(int i = 1; i < N - 1; ++i) {
        if(A[i] < 0) {
            ans.emplace_back(A[N - 1], A[i]);
            A[N - 1] -= A[i];
        } else {
            ans.emplace_back(A[0], A[i]);
            A[0] -= A[i];
        }
    }
    ans.emplace_back(A[N - 1], A[0]);
    cout << A[N - 1] - A[0] << '\n';
    for(auto &[x, y] : ans) cout << x << " " << y << '\n';
    return 0;
}
