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>
#include <vector>
#include <algorithm>
using namespace std;
int main(){
  int n;
  cin >> n;
  vector<int> as;
  for(int i = 0; i < n; i++){
    int a;
    cin >> a;
    as.push_back(a);
  }
  auto minIx = min_element(as.begin(), as.end());
  auto maxIx = max_element(as.begin(), as.end());

  min_element(as.begin(), as.end());
  auto printIter = abs(*minIx) < abs(*maxIx)
               ? maxIx
               : minIx;
  auto printIx = distance(as.begin(), printIter) + 1;
  if( abs(*minIx) < abs(*maxIx) ){
    cout << n*2-1 << endl;
    for(int i = 1; i <= n; i++) {
      cout << printIx << " " << i << endl;
    }

    for(int i = 1; i < n; i++){
      cout << i << " " << i + 1 << endl;
    }
    
  } else {
    cout << n*2-1 << endl;
    for(int i = 1; i <= n; i++) {
      cout << printIx << " " << i << endl;
    }
    for(int i = n; i > 1; i--){
      cout << i << " " << i - 1 << endl;
    }
  }
}
