Problem: 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

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#define MOD 1000000007
#define INF 1 << 30

using namespace std;
typedef long long ll;

int main(void) {
	int num, i, ma = 1, mi = 1, x;
	cin >> num;
	cout << 2 * num - 1 << "\n";
	vector<int> d(num + 1);
	for (i = 1; i <= num; i++) {
		cin >> d[i];
		if (d[ma] < d[i])
			ma = i;
		if (d[mi] > d[i])
			mi = i;
	}
	if (abs(d[ma]) >= abs(d[mi]))
		x = ma;
	else
		x = mi;
	for (i = 1; i <= num; i++)
		printf("%d %d\n", x, i);
	if (d[x] > 0) {
		for(i = 1; i< num; i++)
			printf("%d %d\n", i, i + 1);
	}
	else {
		for (i = num; i > 1; i--)
			printf("%d %d\n", i, i - 1);
	}
	return 0;
}