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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;

int main(){
  vector<int> v,u;
  int N,i,absm=0,p=0,flag=0,count=0,M=2e-6,m=2e6;
  cin >> N;
  int a[N];
  for(i=0;i<N;i++){
    cin >> a[i];
    if(a[i]>M)M=a[i];
    if(a[i]<m)m=a[i];
  }
  if(m<0 && M>=0){
    if(abs(M)-abs(m)>=0)absm = M;
    else absm = m;
  }
  for(i=0;i<N;i++){
    if(a[i]==absm)continue;
    else {
      a[i]+=absm;
      v.push_back(p+1);
      u.push_back(i+1);
      count++;
    }
  }
  //for(i=0;i<N;i++)cout << a[i] << endl;//
  for(i=0;i<N-1;i++){
    if(a[i]<=a[i+1])continue;
    else{
        a[i+1]+=a[i];
        v.push_back(i+1);
        u.push_back(i+2);
        count++;
    }
  }
  cout << count << endl;
  if(!(v.empty())){
    for(i=0;i<v.size();i++){
      cout << v[i] << " " << u[i] << endl;
    }
  }
  //for(i=0;i<N;i++)cout << a[i] << endl;
  return 0;
}