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;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; i++)
#define all(a) (a).begin(), (a).end()

int N;
vector<int> a;

int main () {
  cin >> N;
  a.resize(N);
  rep(i,N) cin >> a[i];

  int iMax = distance(a.begin(), max_element(all(a)));
  int iMin = distance(a.begin(), min_element(all(a)));

  cout << 2*N - 1 << endl;
  if(abs(a[iMax]) > abs(a[iMin])) {
    rep(i,N) if(i != iMax) cout << iMax << " " << i << endl;
  } else {
    rep(i,N) if(i != iMin) cout << iMin << " " << i << endl;
  }

  rep(i,N-1) {
    if(a[0] >= 0) {
      cout << i << " " << i+1 << endl;
    } else {
      cout << N-1-i << " " <<  N-2-i;
    }
  }

  return 0;
}