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;
#define ll long long
typedef pair<int,int> pi;
#define f first
#define s second
#define FAST ios_base::sync_with_stdio(0); cin.tie(0);

int n;
int main() {
	FAST
	cin >> n;
	vector <int> a(n);
	vector <pi> ans;
	for (int i =0;i<n;i++) cin >> a[i];
	auto maxn = max_element(a.begin(),a.end());
	auto minn = min_element(a.begin(),a.end());	
	int maxp = maxn - a.begin() + 1, minp = minn - a.begin() + 1;
	
	if (abs(*maxn) >= abs(*minn)) {
		for (int i =1;i<=n;i++) ans.push_back(pi(maxp,i));
		for (int i =2;i<=n;i++) ans.push_back(pi(i-1,i));
		
	} else if (abs(*maxn) < abs(*minn)) {
		for (int i = 1;i<=n;i++) ans.push_back(pi(minp,i));
		for (int i =n-1;i >= 1; i--) ans.push_back(pi(i+1,i));
	}
	cout << ans.size() << "\n";
	for (auto cur: ans) cout << cur.f << " "<< cur.s << "\n";
	
}	

