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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let n: usize = input();
    let a: Vec<i64> = (0..n).map(|_| input()).collect();

    let (min_idx, &min) = a.iter().enumerate().min_by_key(|&(_, &a)| a).unwrap();
    let (max_idx, &max) = a.iter().enumerate().max_by_key(|&(_, &a)| a).unwrap();

    let out = std::io::stdout();
    let mut buf = std::io::BufWriter::new(out.lock());

    if max == min {
        writeln!(buf, "{}", 0).unwrap();
    } else if min >= 0 {
        writeln!(buf, "{}", n - 1).unwrap();
        for i in 1..n {
            writeln!(buf, "{} {}", i, i + 1).unwrap();
        }
    } else if max <= 0 {
        writeln!(buf, "{}", n - 1).unwrap();
        for i in (1..n).rev() {
            writeln!(buf, "{} {}", i + 1, i).unwrap();
        }
    } else if min.abs() <= max.abs() {
        writeln!(buf, "{}", n + n - 1).unwrap();
        for i in 1..n + 1 {
            writeln!(buf, "{} {}", max_idx + 1, i).unwrap();
        }
        for i in 1..n {
            writeln!(buf, "{} {}", i, i + 1).unwrap();
        }
    } else {
        writeln!(buf, "{}", n + n - 1).unwrap();
        for i in 1..n + 1 {
            writeln!(buf, "{} {}", min_idx + 1, i).unwrap();
        }
        for i in (1..n).rev() {
            writeln!(buf, "{} {}", i + 1, i).unwrap();
        }
    }
}