#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;
typedef long long int ll;

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int n; cin >> n;
	vector<int> a(n);
	int m=0;
	int id=0;
	for(int i=0;i<n;i++){
		cin >> a[i];
		if(abs(m)<abs(a[i])){
			id=i;
			m=a[i];
		}
	}
	cout << 2*n-1 << endl;
	for(int i=0;i<n;i++){
		cout << id+1 << " " << i+1 << endl;
	}
	if(m>0){
		for(int i=0;i<n-1;i++){
			cout << i+1 << " " << i+2 << endl;
		}
	}
	else{
		for(int i=0;i<n-1;i++){
			cout << n-i << " " << n-1-i << endl;
		}
	}
}

### Prompt

Your task is to create 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 <vector>
#include <algorithm>
#include <map>
using namespace std;
typedef long long int ll;

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int n; cin >> n;
	vector<int> a(n);
	int m=0;
	int id=0;
	for(int i=0;i<n;i++){
		cin >> a[i];
		if(abs(m)<abs(a[i])){
			id=i;
			m=a[i];
		}
	}
	cout << 2*n-1 << endl;
	for(int i=0;i<n;i++){
		cout << id+1 << " " << i+1 << endl;
	}
	if(m>0){
		for(int i=0;i<n-1;i++){
			cout << i+1 << " " << i+2 << endl;
		}
	}
	else{
		for(int i=0;i<n-1;i++){
			cout << n-i << " " << n-1-i << endl;
		}
	}
}

```