Problem:
Score : 500 points

Problem Statement

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}

Sample Input 1

3
1 -1 2

Sample Output 1

4
-1 1
2 -2

If we choose x = -1 and y = 1 in the first operation, the set of integers written on the blackboard becomes (-2, 2).

Then, if we choose x = 2 and y = -2 in the second operation, the set of integers written on the blackboard becomes (4).

In this case, we have 4 as the final integer. We cannot end with a greater integer, so the answer is 4.

Sample Input 2

3
1 1 1

Sample Output 2

1
1 1
1 0

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

void merge_sort(int x[], int n)
{
	static int y[100001] = {};
	if (n == 1) return;
	merge_sort(&(x[0]), n/2);
	merge_sort(&(x[n/2]), (n+1)/2);
	
	int i, p, q;
	for (i = 0, p = 0, q = n/2; i < n; i++) {
		if (p >= n/2) y[i] = x[q++];
		else if (q >= n) y[i] = x[p++];
		else y[i] = (x[p] < x[q])? x[p++]: x[q++];
	}
	for (i = 0; i < n; i++) x[i] = y[i];
}

int main()
{
	int i, N, A[100001];
	scanf("%d", &N);
	for (i = 0; i < N; i++) scanf("%d", &(A[i]));
	merge_sort(A, N);
	
	int l, r, tmp[100001];
	if (N == 2) {
		printf("%d\n%d %d\n", A[1] - A[0], A[1], A[0]);
	} else if (A[0] >= 0) {
		tmp[0] = A[0];
		tmp[1] = A[0] - A[N-1];
		for (i = 2; i < N - 1; i++) tmp[i] = tmp[i-1] - A[i-1];
		tmp[N-1] = A[N-2] - tmp[N-2];
		printf("%d\n%d %d\n", tmp[N-1], A[0], A[N-1]);
		for (i = 2; i < N - 1; i++) printf("%d %d\n", tmp[i-1], A[i-1]);
		printf("%d %d\n", A[N-2], tmp[N-2]);
	} else if (A[N-1] <= 0) {
		tmp[0] = A[N-1];
		tmp[1] = A[N-1] - A[0];
		for (i = 2; i < N - 1; i++) tmp[i] = tmp[i-1] - A[i-1];
		tmp[N-1] = tmp[N-2] - A[N-2];
		printf("%d\n%d %d\n", tmp[N-1], A[N-1], A[0]);
		for (i = 2; i < N - 1; i++) printf("%d %d\n", tmp[i-1], A[i-1]);
		printf("%d %d\n", tmp[N-2], A[N-2]);
	} else {
		tmp[0] = A[N-1];
		for (l = 1; A[l] <= 0; l++) tmp[l] = tmp[l-1] - A[l];
		tmp[l] = A[0];
		for (r = N - 2, i = l + 1; A[r] > 0; r--, i++) tmp[i] = tmp[i-1] - A[r];
		tmp[N] = tmp[l-1] - tmp[N-1];
		printf("%d\n", tmp[N]);
		for (i = 1; i < l; i++) printf("%d %d\n", tmp[i-1], A[i]);
		for (r = N - 2, i = l + 1; i < N; r--, i++) printf("%d %d\n", tmp[i-1], A[r]);
		printf("%d %d\n", tmp[l-1], tmp[N-1]);
	}
	fflush(stdout);
	return 0;
}