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;
using pii = pair<int, int>;
int main() {
	int N;
	cin >> N;
	vector<int> a(N);
	vector<pii> ans;
	int M = 0;
	int Mi = -1;
	for(int i = 0; i < N; i++) {
		cin >> a[i];
		if(abs(M) < abs(a[i])) {
			M = a[i];
			Mi = i;
		}
	}
	if(M == 0) {
		cout << 0 << endl;
		return 0;
	}
	for(int i = 0; i < N; i++) {
		if(Mi == i) continue;
		a[i] += M;
		ans.emplace_back(Mi + 1, i + 1);
	}
	if(M > 0) {
		for(int i = 1; i < N; i++) {
			ans.emplace_back(i, i + 1);
			a[i] += a[i - 1];
		}
	} else {
		for(int i = N - 1; i > 0; i--) {
			ans.emplace_back(i + 1, i);
			a[i - 1] += a[i];
		}
	}
	cout << ans.size() << endl;
	for(auto it = ans.begin(); it != ans.end(); it++)
		cout << (*it).first << " " << (*it).second << endl;
	return 0;
}