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
#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

int main() {
	int N;
	cin >> N;

	vector<int> A(N);

	for (int &a : A) {
		cin >> a;
	}
	
	int a_max, a_min,idx_max,idx_min;

	idx_max = idx_min = 0;
	a_max = a_min = A[0];
	for (int i = 1; i < N; i++) {
		if (a_max < A[i]) {
			idx_max = i;
			a_max = A[i];
		}
		if (a_min > A[i]) {
			idx_min = i;
			a_min = A[i];
		}
	}

	if (a_max == a_min) {
		cout << 0 << endl;
	}
	else {
		cout << 2*N - 2 << endl;
		if (abs(a_max) >= abs(a_min)) {
			for (int i = 0; i < N; i++)
			{
				if (i != idx_max) {
					cout << idx_max + 1 << " " << i+ 1 << endl;
				}
			}
			for (int i = 1; i < N; i++) {
				cout << i << " " << i + 1 << endl;
			}
		}
		else {
			for (int i = 0; i < N; i++)
			{
				if (i != idx_min) {
					cout << idx_min + 1 << " " << i + 1 << endl;
				}
			}
			for (int i = N; i > 1; i--) {
				cout << i << " " << i - 1 << endl;
			}
		}
	}


	return 0;
}