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>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include<algorithm>
#include<sstream>
#include<iomanip>


using namespace std;

typedef long long ll;

const ll MOD_CONST = 1000000007;
const ll BIG_NUM = 1000000000000000000;



int main() {
	int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n;i++) {
		cin >> a[i];
	}

	int min_i = 0;
	int max_i = 0;

	for (int i = 1; i < n;i++) {
		if (a[i] < a[min_i]) {
			min_i = i;
		}
		if (a[i] >a[max_i]) {
			max_i = i;
		}
	}
	
	cout << 2 * n -1<< endl;
	if ((-1*a[min_i]) <= (a[max_i])) {
		max_i++;
		for (int i = 1; i <=n;i++) {
			cout << max_i << " " << i << endl;
		}
		for (int i = 1; i < n;i++) {
			cout << i << " " << i+1 << endl;
		}
	}
	else {
		min_i++;
		for (int i = 1; i <= n;i++) {
			cout << min_i << " " << i << endl;
		}
		for (int i = n; i >1;i--) {
			cout << i << " " << i - 1 << endl;
		}
	}
}
	
