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 OCaml (4.02.3) (judge verdict: Wrong Answer)

Scanf.scanf "%d" (fun n ->
    let a = Array.init n (fun _ -> Scanf.scanf " %d" (fun a -> a)) in

    let rec loop i m mi =
        if i = n then mi else
            if abs a.(i) > abs m then loop (i + 1) a.(i) i
                                 else loop (i + 1) m mi
    in
    let mi = loop 0 0 0 in
    let m = a.(mi) in
    let adjust = Array.fold_left (fun acc v -> if v * m < 0 then acc + 1 else acc) 0 a in
    Printf.printf "%d\n" (adjust + n - 1);
    Array.iteri (fun i v ->
        if v * m < 0 then Printf.printf "%d %d\n" (mi + 1) (i + 1)) a;
    if m > 0 then for i = 0 to n - 2     do Printf.printf "%d %d\n" i (i + 1) done
             else for i = n - 1 downto 1 do Printf.printf "%d %d\n" i (i - 1) done

)