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>
#include <math.h>
using namespace std;

int main() {
  int N,X,Y,Z,A,B,C,ans,b,max,min;
  cin >>N;
  int a[N];
  max=-100001;
  min=1000001;
  for(int i=0;i<N;i++){
  cin>>a[i];
  if(a[i]>max){
  max=a[i];
    C=i;
  }
      if(a[i]<min){
  min=a[i];
    A=i;
  }
  }
  if(abs(max)>abs(min)){
    cout<<2*N-1<<endl;
    for(int i=0;i<N;i++){
      cout<<C+1<<" "<<i+1<<endl;
    }
    if(a[C]<=0){
      for(int i=N-1;i>0;i--){
        cout<<i+1<<" "<<i<<endl;
      }
      return 0;
    }
    else{
            for(int i=0;i<N-1;i++){
        cout<<i+1<<" "<<i+2<<endl;
      }
      return 0;
    }
  }
  else {
    cout<<2*N-1<<endl;
    for(int i=0;i<N;i++){
      cout<<A+1<<" "<<i+1<<endl;
    }
        if(a[A]<=0){
      for(int i=N-1;i>0;i--){
        cout<<i+1<<" "<<i<<endl;
      }
      return 0;
    }
    else{
            for(int i=0;i<N-1;i++){
        cout<<i+1<<" "<<i+2<<endl;
      }
      return 0;
    }
  }
}
