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<bits/stdc++.h>
using namespace std;

const int N = 50;
int arr[N + 5];

int main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n;
	cin>>n;
	int posmax = 0, negmax = 0;
	for(int i = 1;i <= n;i++) {
		cin>>arr[i];
		if(arr[i] >= 0) posmax = max(posmax, arr[i]);
		else negmax = max(negmax, -arr[i]);
	}
	int target = posmax;
	if(posmax <= negmax) target = -negmax;
	int idx = 1;
	for(int i = 1;i <= n;i++) {
		if(target == arr[i]) idx = i;
	}
	vector<pair<int,int> > res;
	for(int i = 1;i <= n;i++) {
		arr[i] += arr[idx];
		res.push_back({idx, i});
	}
	if(arr[1] >= 0) {
		for(int i = 2;i <= n;i++) {
			arr[i] += arr[i - 1];
			res.push_back({i - 1, i});
		}
	} else {
		for(int i = n - 1;i >= 1;i--) {
			arr[i] += arr[i + 1];
			res.push_back({i + 1, i});
		}		
	}
	cout<<res.size()<<endl;
	for(int i = 0;i < res.size();i++) cout<<res[i].first<<" "<<res[i].second<<endl;
}