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;
#define rep(i,n) for (int i=0; i<(int)(n); ++i)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using Graph = vector<vi>;  // vi <=> vl
using vb = vector<bool>;
using boolGraph = vector<vb>;
using P = pair<int, int>;  // int <=> ll
const int MOD = 1e9+7;
const double EPS = 1e-9;

int main() {
  int N; cin >> N;
  vi arr(N);
  rep(i,N) cin >> arr[i];

  cout << 2 * (N-1) << endl;

  int MAX = 0, MIN = 0;
  rep(i,N) {
    if (arr[i] > arr[MAX]) MAX = i;
    if (arr[i] < arr[MIN]) MIN = i;
  }

  if (abs(arr[MAX]) > abs(arr[MIN])) {
    rep(i,2) cout << MAX+1 << ' ' << 2 << endl;
    for (int i=2; i<N; ++i) {
      rep(j,2) cout << i << ' ' << i+1 << endl;
    }
  }
  else {
    rep(i,2) cout << MIN+1 << ' ' << N-1 << endl;
    for (int i=N-1; i>0; --i) {
      rep(j,2) cout << i << ' ' << i-1 << endl;
    }
  }
  return 0;
}