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::cmp;
use std::io;


fn read_line() -> String {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line).unwrap();
    return String::from(line.trim());
}

fn main() {
    let _ = read_line().parse::<i64>().unwrap();
    let values: Vec<i64> = read_line().split_whitespace().map(|v| v.parse::<i64>().unwrap()).collect();
    let min = *(values.iter().min().unwrap());
    let max = *(values.iter().max().unwrap());
    let min_index = values.iter().enumerate().filter(|&(_, v)| *v == min).nth(0).unwrap().0;
    let max_index = values.iter().enumerate().filter(|&(_, v)| *v == max).nth(0).unwrap().0;

    let p = |i0, i1| {println!("{} {}", i0 + 1, i1 + 1)};

    println!("{}", (values.len() - 1) * 2);
    if (0 < min) || (-min < max) {
        for i in (0..values.len()).filter(|i| *i != max_index) {
            p(max_index, i);
        }
        for i in 0..values.len() - 1 {
            p(i, i + 1);
        }
    } else {
        for i in (0..values.len()).filter(|i| *i != min_index) {
            p(min_index, i);
        }
        for i in (0..values.len() - 1).rev() {
            p(i + 1, i);
        }
    }
}