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 <string>

using namespace std;

inline int length(int num) {
	return num >= 0 ? num : -num;
}

int main()
{
	/*
		まず全入力を取得。その際最大値と最小値のうち絶対値が大きい方を保持。
		それが最大値ならindex=0から並べていく。逆ならindex = n-1から。
		二回足したら最大値をそちらにシフト。
	
	*/
	int n;
	cin >> n;
	int longest = 0;
	int maxIndex = 1;
	int num;
	for (int i = 1; i <= n; i++) {
		cin >> num;
		if (length(longest) < length(num)) {
			longest = num;
			maxIndex = i;
		}
	}

	int start;
	int end;
	int dir;

	if (longest > 0) {
		start = 2;
		end = n;
		dir = 1;
	}
	else {
		start = n-1;
		end = 1;
		dir = -1;
	}
	string ans = to_string(2 * (n - 1)).append("\n");
	string str = to_string(maxIndex).append(" ").append(to_string(start)).append("\n");
	ans.append(str).append(str);

	for (int i = start + dir; i != end + dir; i += dir) {
		str = to_string(i - dir).append(" ").append(to_string(i)).append("\n");
		ans.append(str).append(str);
	}
	cout << ans;
}