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 (1.42.0) (judge verdict: Wrong Answer)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::collections::*;

fn solve() {
    input! {
        n: usize,
        av: [isize; n]
    };

    let mut min = std::isize::MAX;
    let mut min_i = 0;
    for i in 0..n {
        let a = av[i];
        if a < min {
            min = a;
            min_i = i;
        }
    }
    let mut max = std::isize::MIN;
    let mut max_i = 0;
    for i in 0..n {
        let a = av[i];
        if a > max {
            max = a;
            max_i = i;
        }
    }
    if min.abs() > max.abs() {
        println!("{} {}", min_i, n - 1);
        println!("{} {}", min_i, n - 1);
        for i in (0..n - 1).rev() {
            if av[i] > 0 {
                println!("{} {}", i + 1, i);
                println!("{} {}", i + 1, i);
            } else {
                println!("{} {}", i + 1, i);
            }
        }
    } else {
        println!("{} {}", max_i, 0);
        println!("{} {}", max_i, 0);
        for i in 1..n {
            if av[i] > 0 {
                println!("{} {}", i - 1, i);
            } else {
                println!("{} {}", i - 1, i);
                println!("{} {}", i - 1, i);
            }
        }
    }
}

fn main() {
    std::thread::Builder::new()
        .name("big stack size".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(|| {
            solve();
        })
        .unwrap()
        .join()
        .unwrap();
}