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;

int n;
vector<int> a;
int now = 0;
vector<pair<int, int>> ans;

void solve();

int main() {
  cin >> n;
  for(int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    a.push_back(x);
  }
  solve();
  cout << ans.size() << endl;
  for(int i = 0; i < ans.size(); ++i) {
    cout << ans[i].first + 1 << " " << ans[i].second + 1
         << endl;
  }
  return 0;
}

void solve() {
  while(now < a.size() - 1) {
    if(a[now] <= a[now + 1]) {
      ++now;
      continue;
    }
    else {
      if(a[now] > 0 && a[now + 1] > 0) {
        // example: 5 4
        if(a[now] <= a[now + 1] * 2) {
          ans.push_back(make_pair(now + 1, now + 1));
          a[now + 1] *= 2;
        }
        // example: 10 4
        else {
          ans.push_back(make_pair(now, now + 1));
          a[now + 1] += a[now];
        }
      }
      else if(a[now] <= 0 && a[now + 1] < 0) {
        // example -4 -6
        if(a[now] * 2 <= a[now + 1]) {
          ans.push_back(make_pair(now, now));
          a[now] *= 2;
        }
        // example -2 -6
        else {
          ans.push_back(make_pair(now + 1, now));
          a[now] += a[now + 1];
        }
      }
      // example 5 -10
      else {
        if(a[now] > a[now + 1]) {
          ans.push_back(make_pair(now, now + 1));
          a[now + 1] += a[now];
        }
        else {
          ans.push_back(make_pair(now + 1, now));
          a[now] += a[now + 1];
        }
      }
    }
  }
}