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.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = Integer.parseInt(sc.next());
		long[] a = new long[n];
		long sum = 0;
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(sc.next());
			sum += a[i];
		}
		String[] ans = new String[2 * n];
		int c  = 0;
		if (sum >= 0) {
			for (int i = 1; i < n; i++) {
				int index = 0;
				long max = Integer.MIN_VALUE;
				for (int j = 0; j < n; j++) {
					if (a[j] > max) {
						max = a[j];
						index = j;
					}
				}
				while (a[i] < a[i - 1]) {
					a[i] += max;
					ans[c++] = String.valueOf(index + 1) + " " + String.valueOf(i + 1);
				}
			}
		}
		else {
			for (int i = n - 2; i >= 0; i--) {
				int index = 0;
				long min = Integer.MAX_VALUE;
				for (int j = 0; j < n; j++) {
					if (a[j] < min) {
						min = a[j];
						index = j;
					}
				}
				while (a[i] > a[i + 1]) {
					a[i] += min;
					ans[c++] = String.valueOf(index + 1) + " " + String.valueOf(i + 1);
				}
			}
		}
		System.out.println(c);
		for (int i = 0; i < c; i++)
			System.out.println(ans[i]);
	}
}