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<cstdio>
#include<cmath>
using namespace std;
int main(){
  int n,a[50],mina,minp,maxa,maxp;
  scanf("%d",&n);
  for(int i=0;i<n;i++)
    scanf("%d",&a[i]);
  maxa=mina=a[0];
  minp=maxp=0;
  for(int i=1;i<n;i++){
    if(a[i]<mina){
      mina=a[i];
      minp=i;
    }
    if(a[i]>maxa){
      maxa=a[i];
      maxp=i;
    }
  }
  printf("%d\n",2*n-1);
  for(int i=1;i<=n;i++){
    if(abs(mina)>abs(maxa))
      printf("%d %d\n",minp+1,i);
    else
      printf("%d %d\n",maxp+1,i);
  }
  if(abs(mina)>abs(maxa))
    for(int i=n;i>1;i--)
      printf("%d %d\n",i,i-1);
  else
    for(int i=1;i<n;i++)
      printf("%d %d\n",i,i+1);
}