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::*;

fn read<T: std::str::FromStr>() -> T {
    let stdin = stdin();
    let mut buf = String::new();
	stdin.lock().read_line(&mut buf);
	buf.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
	read::<String>().trim().split_whitespace()
        .map(|w| w.parse().ok().unwrap()).collect()
}

fn main() {
    let n = read::<usize>();
    let v = read_vec::<i64>();
    let mut mi = 0;
    let mut max = 0;
    for i in 0..n {
        if v[i].abs() > max {
            mi = i;
            max = v[i].abs();
        }
    }
    if max == 0 {
        println!("0");
    } else {
        println!("{}", 2 * n - 1);
        for i in 0..n {
            println!("{} {}",mi+1,i+1);
        }
        if v[mi] > 0 {
            for i in 1..n {
                println!("{} {}",i,i+1);
            }
        } else {
            for i in 1..n {
                println!("{} {}",n-i+1,n-i);
            }
        }
    }
}