#include<iostream>
#include<algorithm>
using namespace std;

int main() {
	int a[55], n, maxi = 0, cur = 0; cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		if (abs(a[i]) > maxi) {
			maxi = abs(a[i]);
			cur = i;
		}
	}
	cout << n * 2 - 1 << endl;
	for (int i = 0; i < n; i++)cout << cur + 1 << ' ' << i + 1 << endl;
	if (a[cur] >= 0)for (int i = 0; i < n - 1; i++)cout << i + 1 << ' ' << i + 2 << endl;
	else for (int i = n - 1; i > 0; i--)cout << i + 1 << ' ' << i << endl;

	return 0;
}
### Prompt

Please formulate a CPP solution to the following problem:

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

### Response

```cpp
#include<iostream>
#include<algorithm>
using namespace std;

int main() {
	int a[55], n, maxi = 0, cur = 0; cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		if (abs(a[i]) > maxi) {
			maxi = abs(a[i]);
			cur = i;
		}
	}
	cout << n * 2 - 1 << endl;
	for (int i = 0; i < n; i++)cout << cur + 1 << ' ' << i + 1 << endl;
	if (a[cur] >= 0)for (int i = 0; i < n - 1; i++)cout << i + 1 << ' ' << i + 2 << endl;
	else for (int i = n - 1; i > 0; i--)cout << i + 1 << ' ' << i << endl;

	return 0;
}
```