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 <bits/stdc++.h>

using namespace std;

using int64 = long long;
const int64 INF = 1LL << 60;

int main()
{
  int64 N, A[50];

  cin >> N;
  int64 low = INF, high = -INF;
  for(int i = 0; i < N; i++) {
    cin >> A[i];
    low = min(low, A[i]);
    high = max(high, A[i]);
  }

  vector< pair< int, int > > op;
  if(low < high) {
    int idx = 0;
    for(int i = 0; i < N; i++) {
      if(high == A[i]) idx = i;
    }
    for(int i = 0; i < N; i++) {
      op.emplace_back(idx, i);
    }
    for(int i = 1; i < N; i++) {
      op.emplace_back(i - 1, i);
    }
  } else {
    int idx = 0;
    for(int i = 0; i < N; i++) {
      if(low == A[i]) idx = i;
    }
    for(int i = 0; i < N; i++) {
      op.emplace_back(idx, i);
    }
    for(int i = N - 2; i >= 0; i--) {
      op.emplace_back(i - 1, i);
    }
  }

  cout << op.size() << endl;
  for(auto &p : op) cout << p.first + 1 << " " << p.second + 1 << endl;


}