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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define repr(i, n) for (int i = (n); i >= 0; --i)
#define FOR(i, m, n) for (int i = (m); i < (n); ++i)
#define FORR(i, m, n) for (int i = (m); i >= (n); --i)
#define equals(a, b) (fabs((a) - (b)) < EPS)
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
const ll mod2 = 998244353;
const ll INF = 1e18;
const long double EPS = 1e-10;

int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  rep(i, n) cin >> a[i];

  int maxa = 0, mina = 0;
  rep(i, n) {
    if (a[i] >= 0) maxa = max(maxa, a[i]);
    else mina = min(mina, a[i]);
  }

  int res;
  if (maxa >= -mina) res = maxa;
  else res = mina;
  int idx;
  rep(i, n) {
    if (a[i] == res) {
      idx = i;
      break;
    }
  }

  int m = 2 * n - 1;
  cout << m << endl;

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






  
  return 0;
}