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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (long long i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll MOD = 1000000007;
const ll INF = 1000000000000000000;
#define all(v) v.begin(), v.end()


int main(){
  ll N,ma=-INF,x=0,mi=INF,y=0;
  cin >> N;
  vector<ll> A(N);
  rep(i,N){
    cin >> A.at(i);
    if(A.at(i)>ma){
      ma=max(ma,A.at(i));
      x=i;
    }
    if(mi>A.at(i)){
      mi=min(mi,A.at(i));
      y=i;
    }
  }
  if(abs(ma)<abs(mi)){
    rep(i,N){
      cout << y+1 << " " << i+1 << endl;;
    }
    rep(i,N-1){
      cout << N-i << " " << N-1-i << endl;
    }
  }else{
    rep(i,N){
      cout << x+1 << " " << i+1 << endl;;
    }
    rep(i,N-1){
      cout << i+1 << " " << i+2 << endl;
    }
  }
}