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<bits/stdc++.h>
#define ll long long
using namespace std;
int n,i,mn,mx,a[101],opmx,opmn;
int main(){
	mn=1e9;mx=-1e9;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
      scanf("%d",&a[i]);
      if(a[i]<mn)mn=a[i],opmn=i;
      if(a[i]>mx)mx=a[i],opmx=i;
}
    printf("%d\n",n*2-2);
   if(abs(mx)>=abs(mn)){
   	printf("%d %d\n",opmx,n);
   	printf("%d %d\n",opmx,n);
   	for(i=2;i<n;i++){
   		printf("%d %d\n",n,i);
   		printf("%d %d\n",n,n);
	   }
   }
    else{
    	printf("%d %d\n",opmn,1);
    	printf("%d %d\n",opmn,1);
    	for(i=n-1;i>1;i--){
    		printf("%d %d\n",1,i);
    		printf("%d %d\n",1,1);
		}
	}
}