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;
typedef long long ll;
typedef pair<int,int> P;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	ll N; cin >> N;

	ll a[100];
	ll max_abs = 0;
	ll max_idx = -1;
	for (int i = 0;i < N;i++){
		cin >> a[i];
		if(abs(a[i]) > abs(max_abs)){
			max_abs = a[i];
			max_idx = i;
		}
	}

	vector<P> ans;

	if(max_abs < 0){
		for (int i = N-2;i >= 0;i--){
			while(a[i] > a[i+1]){
				a[i] += max_abs;
				ans.push_back(P(max_idx+1, i+1));
				if(abs(a[i]) > abs(max_abs)){
					max_abs = a[i];
					max_idx = i;
				}
			}
		}
	}else{
		for (int i = 1;i < N;i++){
			while(a[i] < a[i-1]){
				a[i] += max_abs;
				ans.push_back(P(max_idx+1, i+1));
				if(abs(a[i] > abs(max_abs))){
					max_abs = a[i];
					max_idx = i;
				}
			}
		}
	}

	cout << ans.size() << endl;

	for (int i = 0;i < ans.size(); i++){
		cout << ans[i].first << " " << ans[i].second << endl;
	}

	return 0;

}
