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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define rep(i,j,n) for(int i=j;i<n;++i)
#define all(i) i.begin(),i.end()
#define rall(i) i.rbegin(), i.rend()
#define INF 1e9
#define LINF 1e18
const int mod = 1e9 + 7;

typedef long long i64;
typedef pair<int, int> pi;

template <class T> using vt = vector<T>;
template <class T> using vvt = vector<vector<T>>;

i64 gcd(i64 n, i64 m) {return (m == 0? n : gcd(m, n % m));}
i64 lcd(i64 n, i64 m) {return (n / gcd(n, m) * m);}
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;
  vt<int> a(n);
  rep(i, 0, n) cin >> a[i];

  int MAX = max_element(all(a)) - a.begin();
  int MIN = min_element(all(a)) - a.begin();
  if(MAX == MIN) return cout << 0 << endl, 0;

  cout << 2 * n - 1 << endl;
  if(abs(a[MAX]) > abs(a[MIN])) {
    rep(i, 0, n) {
      cout << MAX + 1 << " " << i + 1 << endl;
    }
    rep(i, 1, n) {
      cout << i << " " << i + 1 << endl;
    }
  }
  else {
    rep(i, 0, n) {
      cout << MIN + 1 << " " << i + 1 << endl;
    }
    for(int i = n - 1; i > 0; --i) {
      cout << i + 1 << " " << i << endl;
    }
  }
}