#include<bits/stdc++.h>
using namespace std;
using ll = long long;
int main(){
	ll n; cin >> n;
	ll si = 0, na = 1;
	bool f = 0;
	for(ll i = 0; i < n; ++i){
		ll a; cin >> a;
		if((ll)abs(a) > (ll)abs(si)){
			si = a;
			na = i + 1;
			f = a > 0;
		}
	}
	cout << 2 * n << endl;
	cout << na << " " << (f ? 1 : n) << endl;
	cout << na << " " << (f ? 1 : n) << endl;
	for(ll i = 1; i < n; ++i){
		cout << (f ? i : n + 1 - i) << " " << (f ? i + 1 : n - i) << endl;
		cout << (f ? i : n + 1 - i) << " " << (f ? i + 1 : n - i) << endl;
	}
	return 0;
}
### 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>
using namespace std;
using ll = long long;
int main(){
	ll n; cin >> n;
	ll si = 0, na = 1;
	bool f = 0;
	for(ll i = 0; i < n; ++i){
		ll a; cin >> a;
		if((ll)abs(a) > (ll)abs(si)){
			si = a;
			na = i + 1;
			f = a > 0;
		}
	}
	cout << 2 * n << endl;
	cout << na << " " << (f ? 1 : n) << endl;
	cout << na << " " << (f ? 1 : n) << endl;
	for(ll i = 1; i < n; ++i){
		cout << (f ? i : n + 1 - i) << " " << (f ? i + 1 : n - i) << endl;
		cout << (f ? i : n + 1 - i) << " " << (f ? i + 1 : n - i) << endl;
	}
	return 0;
}
```