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 <algorithm>
#include <vector>
#include <string>
#include <cmath>
#include <map>
using namespace std;

#define INF 1e9

int main() {
	int N; cin >> N;
	vector<int>A(N);
	for (int i = 0; i < N; i++)cin >> A[i];

	int minA = 1e9, maxA = -1e9;
	int minA_index, maxA_index;
	for (int i = 0; i < N; i++) {
		if (minA > A[i]) {
			minA = A[i];
			minA_index = i;
		}
		if (maxA < A[i]) {
			maxA = A[i];
			maxA_index = i;
		}
	}

	cout << 2 * N - 1 << endl;

	if (abs(minA) > abs(maxA)) {
		for (int i = 0; i < N; i++) {
			cout << minA_index+1 << " " << i+1 << endl;
		}

		for (int i = N - 1; i > 0; i--) {
			cout << i + 1 << " " << i - 1 + 1 << endl;
		}
	}
	else {
		for (int i = 0; i < N; i++) {
			cout << maxA_index+1 << " " << i+1 << endl;
		}

		for (int i = 1; i < N; i++) {
			cout << i - 1 + 1 << " " << i + 1 << endl;
		}
	}
	
	return 0;
}