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: Accepted)

    #include <iostream>
    #include <vector>
    #include <utility>
    #include <assert.h>
    using namespace std;
     
    int main(){
      int n,i;
      vector<pair<int,int> > v;
      cin >> n;
      long long a[n];
      int t;
      long long mx = 0,cnt=0;
      bool ch;
      for(i=0;i<n;i++){
        cin >> a[i];
        if(abs(a[i])-abs(mx)>0){
          mx = a[i];
          t=i;
        }
      }
      if(mx>=0)ch=true;
      else ch = false;
      for(i=0;i<n;i++){
        a[i]+=mx;
        cnt++;
        v.push_back(make_pair(t+1,i+1));
      }
      if(ch){
      	for(i=1;i<n;i++){
          if(a[i]<=a[i-1]){
            a[i]+=a[i-1];
            v.push_back(make_pair(i,i+1));
            cnt++;
          }
        }
      }else{
        for(i=n-2;i>=0;i--){
            a[i]+=a[i+1];
            v.push_back(make_pair(i+2,i+1));
            cnt++;
        }
      }
      
      cout << cnt << endl;
      for(i=0;i<v.size();i++){
        cout << v[i].first << " " << v[i].second << endl;
      }
      
      /*
      cout << "===========" << endl;
      for (int i = 0; i < n; i++) {
      	cout << a[i] << endl;
      	if (i + 1 < n) {
      		assert(a[i] <= a[i + 1]);
      	}
      }
      */
      /*for(i=0;i<n;i++)cout << a[i] << " ";
      cout << endl;*/
       return 0;
    }