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
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef pair<int,int> pp;

int N;
const int MAX_N = 55;
int a[MAX_N];
vector<pp> s;

int main(){
	ios::sync_with_stdio(false);

	cin >> N;
	for(int i=1;i<=N;i++){
		cin >> a[i];
	}

	int m = min_element(a+1,a+N+1)-a;
	int M = max_element(a+1,a+N+1)-a;

	if(a[M]+a[m] >= 0){
		for(int i=1;i<=N;i++){
			if(a[i]<0){
				s.push_back({M,i});
			}
		}
		for(int i=1;i<=N-1;i++){
			s.push_back({i,i+1});
		}
	}else{
		for(int i=1;i<=N;i++){
			if(a[i]>0){
				s.push_back({m,i});
			}
		}
		for(int i=N;i>1;i--){
			s.push_back({i,i-1});
		}
	}

	cout << s.size() << endl;
	for(auto& i:s){
		cout << i.first << " " << i.second << "\n";
	}
	return 0;
}