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 <cstdio>
#include <algorithm>
using namespace std;

int main(){
  int N;
  scanf("%d",&N);
  int a[N];
  int i;
  int min,max,min_num=0,max_num=0;
  for(i=0;i<N;i++){
    scanf("%d",&a[i]);
    if(i==0){
      max=a[i];
      min=a[i];
    }
    else{
      if(a[i]<min){
	min=a[i];
	min_num=i;
      }
      if(a[i]>max){
	max=a[i];
	max_num=i;
      }
    }
  }
  printf("%d\n",2*N);
  int j;
  if(min+max>0){
    printf("%d %d\n",max_num+1,1);
    printf("%d %d\n",max_num+1,1);
    for(j=0;j<N-1;j++){
      printf("%d %d\n",j+1,j+2);
      printf("%d %d\n",j+1,j+2);
    }
  }
  else{
    printf("%d %d\n",min_num+1,N);
    printf("%d %d\n",min_num+1,N);
    for(j=0;j<N-1;j++){
      printf("%d %d\n",N-j,N-1-j);
      printf("%d %d\n",N-j,N-1-j);
    }
  }
  return 0;
}
