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 Scala (2.11.7) (judge verdict: Wrong Answer)

import scala.io.{StdIn => in}
import scala.collection.{mutable => mu}

object Main extends App {
  val n = in.readInt
  val a = in.readLine.split(" ").map(_.toInt)

  val max = a.max
  val min = a.min

  val buf = StringBuilder.newBuilder
  if(max == 0 && min == 0) buf.append(0).append('\n')
  else if(min >= 0) {
    (1 until n).foreach(i => buf.append(i).append(' ').append(i+1).append('\n'))
  }
  else if(max < 0) {
    (n-1 to 1 by -1).foreach(i => buf.append(i+1).append(' ').append(i).append('\n'))
  } else {
    val m = if(max.abs > min.abs) max else min
    val idx = a.indexWhere(_ == m)
    for(i <- 0 until n if i != idx){
      buf.append(idx+1).append(' ').append(i+1).append('\n')
    }
    val b = a.map(_ + m)
    if(m >= 0)
      (1 until n).foreach(i => buf.append(i).append(' ').append(i+1).append('\n'))
    else
      (n-1 to 1 by -1).foreach(i => buf.append(i+1).append(' ').append(i).append('\n'))
  }

  print(buf)
}