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>
using namespace std;
 
int main(){
  int n, a[55], mxa = 0, mna = 0, i, k;
  scanf("%d", &n);
  for(i = 0; i < n; i++){
    scanf("%d", &a[i]);
    if(a[mxa] < a[i]) mxa = i;
    if(a[mna] > a[i]) mna = i;
  }
  k = a[mxa] + a[mna] > 0 ? mxa : mna;
  
  if(a[mxa] > 0 && a[mna] < 0){
    printf("%d\n", 2*(n-1));
    for(i = 0; i < n; i++){
      if(i == k) continue;
      printf("%d %d\n", k+1, i+1);
    }
  }else printf("%d\n", n-1);
  
  if(a[k] > 0) for(i = 1; i < n; i++) printf("%d %d\n", i, i+1);
  else for(i = n-1; i > 0; i--) printf("%d %d\n", i+1, i);
}