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 <iostream>

const int INF = 1000001;

int main(){
   int N;
   std::cin >> N;
   int a[N];
   int max = -INF;
   int min = INF;
   int mindex, mandex;
   for(int i=0; i<N; ++i){
      std::cin >> a[i];
      if(max<a[i]){
	 max = a[i];
	 mandex = i+1;
      }
      if(min>a[i]){
	 min = a[i];
	 mindex = i+1;
      }
   }
   if(max==min){
      std::cout << 0 << std::endl;
      return 0;
   }else{	
      std::cout << 2*N-1 << std::endl;
   }
   for(int i=0; i<N; ++i){
      if(abs(max)>abs(min)){
	 a[i] += max;
	 std::cout << mandex << ' ' << i+1 << std::endl;
      }else{
	 a[i] += min;
	 std::cout << mindex << ' ' << i+1 << std::endl;
      }
   }

   if(abs(max)>abs(min)){
      for(int i=0; i<N-1; ++i){
	 std::cout << i+1 << ' ' << i+2 << std::endl;
	 a[i+1] += a[i];
      }
   }else{
      for(int i=N-2; i>=0; --i){
	 std::cout << i+2 << ' ' << i+1 << std::endl;
	 a[i] += a[i+1];
      }
   }


   return 0;
}
