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 <utility>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
	bool reved = false;
	int n;
	cin >> n;
	vector<int> a(n);
	
	vector<pair<int,int>> ans;
	
	int Max = -1, pos;
	
	for(int i = 0; n > i; i++){
		cin >> a[i];
		if(abs(a[i]) > Max){
			Max = abs(a[i]);
			pos = i;
		}
	}
	
	if(a[pos] < 0){
		reverse(a.begin(), a.end());
		reved = true;
		for(int i = 0;n > i; i++)
			a[i] *= -1;
		pos = n - pos - 1;
	}
	
	ans.push_back({pos, 0});
	ans.push_back({pos, 0});
	
	for(int i = 0;n - 1 > i; i++){
		ans.push_back({i, i + 1});
		ans.push_back({i, i + 1});
	}
	
	cout << ans.size() << endl;
	for(auto& b : ans){
		if(reved){
			b.first = n - b.first - 1;
			b.second = n - b.second - 1;
		}
		
		cout << b.first + 1 << " " << b.second + 1<< endl;
	}
	
	
	return 0;
}