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 <algorithm>
#include <iostream>
#include <queue>
#include <string>
#include <utility>
using namespace std;

#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define RFOR(i, a, b) for (int i = a; i > b; --i)
#define REP(i, n) for (int i = 0; i < n; ++i)

const int kMaxN = 50;

int n;
int a[kMaxN];
int ans = 0;
int ans_pair[2*kMaxN][2];

int main() {
  cin >> n;
  REP(i, n) { cin >> a[i]; }

  int y = n;
  REP(i, n) {
    if (a[i] > 0) {
      y = i+1;
      break;
    }
  }
  int i_ans_pair = 0;
  while (y < n) {
    if (a[y-1] > a[y]) {
      int x = y-1;
      REP(i, n) {
        if (a[y-1] <= a[y] + a[i]) {
          x = i;
          break;
        }
      }
      a[y] += a[x];
      ++ans;
      ans_pair[i_ans_pair][0] = x;
      ans_pair[i_ans_pair][1] = y;
      ++i_ans_pair;
    } else {
      ++y;
    }
  }

  y = -1;
  RFOR(i, n-1, -1) {
    if (a[i] < 0) {
      y = i-1;
      break;
    }
  }
  while (y > -1) {
    if (a[y] > a[y+1]) {
      int x = y+1;
      RFOR(i, n-1, -1) {
        if (a[y] + a[i] <= a[y+1]) {
          x = i;
          break;
        }
      }
      a[y] += a[x];
      ++ans;
      ans_pair[i_ans_pair][0] = x;
      ans_pair[i_ans_pair][1] = y;
      ++i_ans_pair;
    } else {
      --y;
    }
  }

  cout << ans << endl;
  REP(i, i_ans_pair) {
    cout << ans_pair[i][0]+1 << " " << ans_pair[i][1]+1 << endl;
  }
  return 0;
}