Problem:
Score : 600 points

Problem Statement

Snuke has an integer sequence, a, of length N. The i-th element of a (1-indexed) is a_{i}.

He can perform the following operation any number of times:

Operation: Choose integers x and y between 1 and N (inclusive), and add a_x to a_y.

He would like to perform this operation between 0 and 2N times (inclusive) so that a satisfies the condition below. Show one such sequence of operations.
It can be proved that such a sequence of operations always exists under the constraints in this problem.

Condition: a_1 \leq a_2 \leq ... \leq a_{N}

Constraints

2 \leq N \leq 50

-10^{6} \leq a_i \leq 10^{6}

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{N}

Output

Let m be the number of operations in your solution. In the first line, print m.
In the i-th of the subsequent m lines, print the numbers x and y chosen in the i-th operation, with a space in between.
The output will be considered correct if m is between 0 and 2N (inclusive) and a satisfies the condition after the m operations.

Sample Input 1

3
-2 5 -1

Sample Output 1

2
2 3
3 3

After the first operation, a = (-2,5,4).

After the second operation, a = (-2,5,8), and the condition is now satisfied.

Sample Input 2

2
-1 -3

Sample Output 2

1
2 1

After the first operation, a = (-4,-3) and the condition is now satisfied.

Sample Input 3

5
0 0 0 0 0

Sample Output 3

0

The condition is satisfied already in the beginning.

Solution in Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    var N = readLine()!!.toInt()
    var A = readLine()!!.split(" ").map { it.toInt() }
    when {
    // 非負だけ
        A.filter { it >= 0 }.size == N -> {
            for (i in 1..N - 1) {
                println("$i ${i + 1}")
            }
        }
    // 非正だけ
        A.filter { it <= 0 }.size == N -> {
            for (i in N - 1 downTo 1) {
                println("${i + 1} $i")
            }
        }
    // 混在
        else -> {
            val max = A.withIndex().maxBy { it.value }!!
            val min = A.withIndex().minBy { it.value }!!
            when {
                Math.abs(max.value) >= Math.abs(min.value) -> {
                    for (i in 1..N) {
                        println("${max.index + 1} $i")
                    }
                    for (i in 1..N - 1) {
                        println("$i ${i + 1}")
                    }
                }
                else -> {
                    for (i in 1..N) {
                        println("${min.index + 1} $i")
                    }
                    for (i in N - 1 downTo 1) {
                        println("${i + 1} $i")
                    }
                }
            }
        }
    }
}