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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n];
		int[] min = new int[2];
		int[] max = new int[2];

		min[0] = Integer.MAX_VALUE;
		max[0] = Integer.MIN_VALUE;

		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
			if (min[0] > a[i]) {
				min[0] = a[i];
				min[1] = i;
			}
			if (max[0] < a[i]) {
				max[0] = a[i];
				max[1] = i;
			}
		}

		List<int[]> answer = new ArrayList<>();
		if (min[0] >= 0) {
			for (int i = 1; i < n; i++) {
				answer.add(new int[]{i, i + 1});
			}
		} else if (max[0] <= 0) {
			for (int i = n - 1; i > 0; i--) {
				answer.add(new int[]{i + 1, i});
			}
		} else {
			if (Math.abs(min[0]) > Math.abs(max[0])) {
				for (int i = 0; i < n; i++) {
					if (a[i] > 0) answer.add(new int[]{min[1] + 1, i + 1});
				}
				for (int i = n - 1; i > 0; i--) {
					answer.add(new int[]{i + 1, i});
				}
			} else {
				for (int i = 0; i < n; i++) {
					if (a[i] < 0) answer.add(new int[]{max[1] + 1, i + 1});
				}
				for (int i = 1; i < n; i++) {
					answer.add(new int[]{i, i + 1});
				}
			}
		}

		System.out.println(answer.size());
		for (int[] x : answer) {
			System.out.println(x[0] + " " + x[1]);
		}

	}
}