Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
-2 5 -1

Sample Output 1

2
2 3
3 3

After the first operation, a = (-2,5,4).

After the second operation, a = (-2,5,8), and the condition is now satisfied.

Sample Input 2

2
-1 -3

Sample Output 2

1
2 1

After the first operation, a = (-4,-3) and the condition is now satisfied.

Sample Input 3

5
0 0 0 0 0

Sample Output 3

0

The condition is satisfied already in the beginning.

Solution in JavaScript (node.js v5.12) (judge verdict: Accepted)

function Main(input) {
  input = input.split("\n");
  var n = parseInt(input[0].split(" ")[0]);
  tmp=input[1].split(" ");
  a = new Array(n).fill(0);
  b = new Array();
  var max=-Math.pow(10,6)-1;
  var min=Math.pow(10,6)+1;
  for(i=0;i<n;i++){
    a[i]=parseInt(tmp[i]);
    max=Math.max(a[i],max);
    min=Math.min(a[i],min);
  }
  if(max>=Math.abs(min)){
    for(i=0;i<n;i++){
      if(a.indexOf(max)!=i){
        bsub=[a.indexOf(max)+1,i+1];
        b.push(bsub);
        a[i]=a[i]+max;
      }
    }
    for(i=0;i<n;i++){
      if(i>0&&a[i]<a[i-1]){
        bsub=[i,i+1];
        b.push(bsub);
        a[i]=a[i-1]+a[i];
      }
    }
  }else if(Math.abs(min)>max){
    for(i=0;i<n;i++){
      if(a.indexOf(min)!=i){
        bsub=[a.indexOf(min)+1,i+1];
        b.push(bsub);
        a[i]=a[i]+min;
      }
    }
    for(i=n-1;i>=0;i--){
      if(i<n-1&&a[i]>a[i+1]){
        bsub=[i+2,i+1];
        b.push(bsub);
        a[i]=a[i+1]+a[i];
      }
    }
  }
  var ans =b.length;
  for(i=0;i<b.length;i++){
    ans+="\n"+b[i][0]+" "+b[i][1];
  }
  console.log('%s', ans);
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));