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;

const int MAX_N = 114514;

typedef pair<int, int> pii;

int N, A[MAX_N];
vector<pii> opers;

int main() {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
    sort(A + 1, A + N + 1);
    int split = 1;
    while (split + 1 < N && A[split + 1] < 0) split++;
    int curr = A[1];
    for (int i = N; i > split + 1; i--) {
        opers.emplace_back(curr, A[i]);
        curr -= A[i];
    }
    opers.emplace_back(A[split + 1], curr);
    curr = A[split + 1] - curr;
    for (int i = 2; i <= split; i++) {
        opers.emplace_back(curr, A[i]);
        curr -= A[i];
    }
    printf("%d\n", curr);
    for (auto i : opers) {
        printf("%d %d\n", i.first, i.second);
    }
}