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;

const int N = 2e5;

typedef pair<int,int> pii;

int n,m;
int li[N+5];

vector<pii> com;

int main(){
	scanf("%d",&n);
	int maksi,mini;
	maksi = mini = 1;
	for(int i=1;i<=n;i++){
		scanf("%d",&li[i]);
		if(li[i] > li[maksi]) maksi = i;
		if(li[i] < li[mini]) mini = i;
	}
	if(-li[mini] > li[maksi]){
		for(int i=1;i<=n;i++){
			if(li[i] > 0){
				com.push_back(make_pair(mini,i));
				li[i] += li[mini];
			}
		}
		
		for(int i=n-1;i>=1;i--){
			com.push_back(make_pair(i+1,i));
		}
	} else {
		for(int i=1;i<=n;i++){
			if(li[i] < 0){
				com.push_back(make_pair(maksi,i));
				li[i] += li[mini];
			}
		}
		
		for(int i=2;i<=n;i++){
			com.push_back(make_pair(i-1,i));
		}
	}
	printf("%d\n",com.size());
	for(int i=0;i<com.size();i++){
		printf("%d %d\n",com[i].first,com[i].second);
	}
	return 0;
}