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 JAVA):

Solution:
import java.util.*;

class Main{
    public static void main(String[]args){
        Scanner sc = new Scanner(System.in);
        int iin = sc.nextInt();
        int[] iia = new int[iin];
        String[] sirta = new String[100];
        int idxsirta = 0;
        int imax = -1000001;
        int imaxidx = 0;
        int imin = 1000001;
        int iminidx = 0;
        for(int i=0;i<iin;i++){
          iia[i] = sc.nextInt();
          if(imax < iia[i]){
            imax = iia[i];
            imaxidx = i+1;
          }
          if(imin > iia[i]){
            imin = iia[i];
            iminidx = i+1;
          }
        }
        if(imin>=0){
          for(int i=0;i<iin-1;i++){
            if(iia[i]>iia[i+1]){
              iia[i+1] += iia[i];
              sirta[idxsirta] = (i+1) + " " + (i+2);
              idxsirta++;
            }
          }
        }
        else if(imax<0){
          for(int i=iin-1;i>0;i--){
            if(iia[i-1]>iia[i]){
              iia[i-1] += iia[i];
              sirta[idxsirta] = (i+1) + " " + i;
              idxsirta++;
            }
          }          
        }
        else{
          if(imax-(imin*(-1))>=0){
            for(int i=0;i<iin;i++){
              if(iia[i]<0){
                iia[i] += imax;
                sirta[idxsirta] = imaxidx + " " + (i+1);
                idxsirta++;
              }
            }
            for(int i=0;i<iin-1;i++){
              if(iia[i]>iia[i+1]){
                iia[i+1] += iia[i];
                sirta[idxsirta] = (i+1) + " " + (i+2);
                idxsirta++;
              }
            }
          }
          else{
            for(int i=0;i<iin;i++){
              if(iia[i]>0){
                iia[i] += imin;
                sirta[idxsirta] = iminidx + " " + (i+1);
                idxsirta++;
              }
            }
             for(int i=iin-1;i>0;i--){
              if(iia[i-1]>iia[i]){
                iia[i-1] += iia[i];
                sirta[idxsirta] = (i+1) + " " + i;
                idxsirta++;
              }
            }          
          }
        }
        System.out.println(idxsirta);
        for(int i=0;i<idxsirta;i++){
          System.out.println(sirta[i]);
        }
    }
}