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 <cstdio>
#include <cstring>
#include <climits>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <sstream>
using namespace std;
typedef pair<int, int> pii;
typedef long long int ll;

int N, a;
int m;
int idx;

int main() {

  cin >> N;
  for(int i = 1; i <= N; i++) {
    cin >> a;
    if (abs(a) > abs(m)) {
      m = a;
      idx = i;
    }
  }

  cout << 2 * N - 2 << endl;
  for(int i = 1; i <= N; i++) {
    if (i == idx) {
      continue;
    }
    cout << idx << " " << i << endl;
  }

  for(int i = 1; i < N; i++) {
    if (m >= 0) {
      cout << i << " " << i + 1 << endl;
    } else {
      cout << N - i + 1 << " " << N - i << endl;
    }
  }

  return 0;
}