#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
int main(){
	int N; 
  	cin >> N; 
	int A[60] = {}; 
	for(int i = 0; i < N; i++){
		cin >> A[i]; 
	}
	int absmax = 0; 
	for(int i = 0; i < N; i++){
		if(abs(A[i]) > abs(A[absmax])){
			absmax = i; 
		}
	}
	cout << N * 2 - 1 << endl; 
	for(int i = 0; i < N; i++){
		cout << absmax + 1 << " " << i+1 << endl; 
	}
	if(A[absmax] >= 0){
		for(int i = 0; i < N-1; i++){
			cout << i+1 << " " << i + 2 << endl; 
		}
	}
	if(A[absmax] < 0){
		for(int i = N-2; i >= 0; i--){
			cout << i+2 << " " << i+1 << endl; 
		}
	}
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 <bits/stdc++.h>
#include <algorithm>
using namespace std;
int main(){
	int N; 
  	cin >> N; 
	int A[60] = {}; 
	for(int i = 0; i < N; i++){
		cin >> A[i]; 
	}
	int absmax = 0; 
	for(int i = 0; i < N; i++){
		if(abs(A[i]) > abs(A[absmax])){
			absmax = i; 
		}
	}
	cout << N * 2 - 1 << endl; 
	for(int i = 0; i < N; i++){
		cout << absmax + 1 << " " << i+1 << endl; 
	}
	if(A[absmax] >= 0){
		for(int i = 0; i < N-1; i++){
			cout << i+1 << " " << i + 2 << endl; 
		}
	}
	if(A[absmax] < 0){
		for(int i = N-2; i >= 0; i--){
			cout << i+2 << " " << i+1 << endl; 
		}
	}
}
```