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: Wrong Answer)

// Try AtCoder
// author: Leonardone @ NEETSDKASU

fn main() {
    let mut stdin = String::new();
    std::io::Read::read_to_string(&mut std::io::stdin(), &mut stdin).unwrap();
    let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get { () => ( get().parse().unwrap() ) }
    
    let n = get!();
    let mut vs = vec![0; n];
    let mut p = 0;
    let mut q = 0;
    let mut mi = 0;
    for i in 0..n {
    	vs[i] = get!();
        if vs[i] > vs[p] {
        	p = i;
        }
        if vs[i] < vs[q] {
        	q = i;
        }
        if vs[i] < 0 {
        	mi += 1;
        }
    }
    if mi == 0 {
    	println!("{}", n);
        println!("{} 1", p+1);
        for i in 1..n {
        	println!("{} {}", i, i+1);
        }
        return;
    }
    if mi == n {
    	println!("{}", n);
        println!("{} 1", p+1);
        for i in 0..n-1 {
        	println!("{} {}", n-i, n-i-1);
        }
        return;
    }
    println!("0");
}