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 Ruby (2.3.3) (judge verdict: Wrong Answer)

N, *A = $stdin.read.split.map(&:to_i)

A_MAX = A.max
A_MIN = A.min
B = A.dup

def f(i, n_op)
  if i+1 >= N
    puts n_op
    return true
  end

  if n_op > 2*N
    return false
  end

  if B[i] > B[i+1]
    if A_MAX > 0
      n = (B[i] - B[i+1]) / A_MAX
      n += 1 if (B[i] - B[i+1]) % A_MAX > 0
      B[i+1] += A_MAX * n
      if f(i+1, n_op+n)
        n.times do
          puts "#{A.index(A_MAX)+1} #{i+1+1}"
        end
        return true
      end
      B[i+1] = A[i+1]
    end

    if A_MIN < 0
      n = (B[i] - B[i+1]) / -A_MIN
      n += 1 if (B[i] - B[i+1]) % -A_MIN > 0
      B[i] -= A_MIN * n
      if f(i+1, n_op+n)
        n.times do
          puts "#{A.index(A_MIN)+1} #{i+1}"
        end
        return true
      end
      B[i] = A[i]
    end

  else
    return f(i+1, n_op)
  end

  return false
end

f(0, 0)