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 = gets.strip.to_i
xs = [0] + gets.strip.split.map(&:to_i)
count = 0
ans = []

#p xs
xsmax = xs[1...xs.length].max
xsmin = xs[1...xs.length].min
if xsmax > 0 and xsmin < 0 then
  if xsmax.abs > xsmin.abs then
    ind = xs.index(xsmax)
#    p [xsmax, xs.index(xsmax), ind]

    (1...xs.length).each {|i| xs[i] += xsmax}
    (1...xs.length).each {|i| ans << "#{ind} #{i}"; count += 1}
  else
    ind = xs.index(xsmin)
    (1...xs.length).each {|i| xs[i] += xsmin}
    (1...xs.length).each {|i| ans << "#{ind} #{i}"; count += 1}
  end
end
#p xs
for i in 1...xs.length do
  if xs[i] < xs[i-1] then
    if xs[i] > 0 then
      xs[i] += xs[i-1]
      ans << "#{i-1} #{i}"
      count += 1
    else
      if i > 1 then
        xs[i-1] += xs[i]
        ans << "#{i} #{i-1}"
        count += 1
      end
    end
  end
end
#p xs

puts count
puts ans.join("\n")