#include <iostream>
using namespace std;

int a[55];

int main() {
	int N;
	cin >> N;
	int mx=1;
	for(int i=1; i<=N; ++i){
		cin >> a[i];
		if(abs(a[mx])<abs(a[i])) mx=i;
	}
	cout << 2*(N-1) << endl;
	for(int i=1; i<=N; ++i){
		if(i!=mx) cout << mx << ' ' << i << endl;
	}
	if(a[mx]>0){
		for(int i=1; i+1<=N; ++i){
			cout << i << ' ' << i+1 << endl;
		}
	}else{
		for(int i=N; i>=2; --i){
			cout << i << ' ' << i-1 << endl;
		}
	}
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
using namespace std;

int a[55];

int main() {
	int N;
	cin >> N;
	int mx=1;
	for(int i=1; i<=N; ++i){
		cin >> a[i];
		if(abs(a[mx])<abs(a[i])) mx=i;
	}
	cout << 2*(N-1) << endl;
	for(int i=1; i<=N; ++i){
		if(i!=mx) cout << mx << ' ' << i << endl;
	}
	if(a[mx]>0){
		for(int i=1; i+1<=N; ++i){
			cout << i << ' ' << i+1 << endl;
		}
	}else{
		for(int i=N; i>=2; --i){
			cout << i << ' ' << i-1 << endl;
		}
	}
	return 0;
}
```