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.

Examples

Input

3
-2 5 -1


Output

2
2 3
3 3


Input

2
-1 -3


Output

1
2 1


Input

5
0 0 0 0 0


Output

0
import java.util.*;

public class Main {

	static int n, a[], b[];
	static List<Integer> list[];

	public static void solve(int ptv) {
		if (ptv >= 0) {
			int min = a[0];
			for (int i = 1; i < n; ++i) {
				if (min > a[i]) {
					list[0].add(i);
					list[1].add(i + 1);
					a[i] += min;
				}
				min = a[i];
			}
		} else {
			int max = a[n - 1];
			for (int i = n - 2; i >= 0; --i) {
				if(max < a[i]) {
					list[0].add(i + 2);
					list[1].add(i + 1);
					a[i] += max;
				}
				max = a[i];
			}
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		a = new int[n];
		b = new int[n];
		list = new List[2];
		for (int i = 0; i < 2; ++i)
			list[i] = new ArrayList<Integer>();
		for (int i = 0; i < n; ++i) {
			a[i] = sc.nextInt();
			b[i] = a[i];
		}
		sc.close();
		Arrays.sort(b);
		int pos1 = 0, pos2 = 0;
		for(int i = 0; i < n; ++i) {
			if(a[i] == b[0])pos1 = i;
			if(a[i] == b[n - 1])pos2 = i;
		}
		int pos;
		if (b[0] < 0 && b[n - 1] > 0) {
			int max = b[0];
			pos = pos1;
			if (Math.abs(max) <= b[n - 1]) {
				max = b[n - 1];
				pos = pos2;
			}
			for (int i = 0; i < n; ++i) {
				b[i] += max;
				a[i] += max;
				list[1].add(i + 1);
				list[0].add(pos + 1);
			}
		}
		int ptv = b[0];
		solve(ptv);
		n = list[0].size();
		System.out.println(n);
		for (int i = 0; i < n; ++i) {
			System.out.println(list[0].get(i) + " " + list[1].get(i));
		}
	}

}
