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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
)

func main() {
	var n int
	fmt.Scan(&n)

	scanner := bufio.NewScanner(os.Stdin)
	scanner.Split(bufio.ScanWords)

	a := make([]int, n)
	for i := 0; i < n; i++ {
		scanner.Scan()
		aa, _ := strconv.Atoi(scanner.Text())
		a[i] = aa
	}
	sort.Ints(a)

	var ops [][]int
	var max func(i, j int) int
	var min func(i, j int) int
	max = func(i, j int) int {
		// fmt.Println("max", i, j)
		if i == j {
			return a[i]
		}
		x := a[j]
		y := min(i, j-1)
		ops = append(ops, []int{x, y})
		return x - y
	}
	min = func(i, j int) int {
		// fmt.Println("min", i, j)
		if i == j {
			return a[i]
		}
		x := a[i]
		y := max(i+1, j)
		ops = append(ops, []int{x, y})
		return x - y
	}

	ret := max(0, n-1)
	fmt.Println(ret)
	for i := 0; i < n-1; i++ {
		fmt.Print(ops[i][0])
		fmt.Print(" ")
		fmt.Print(ops[i][1])
		fmt.Println("")
	}
}