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;
const int N = 55;
struct data {
  int x, y;
} d[N * 2];
int a[N];

int main() {
  int n, be = 0, mx = 0;
  scanf("%d", &n);
  int res = 0;
  for(int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if(abs(a[i]) > mx) {
      be = i;
      mx = abs(a[i]);
    }
  }
  if(!be) {
    puts("0");
    exit(0);
  }
  for(int i = 1; i <= n; ++i) {
    if(i != be) {
      d[++res] = (data) {be, i};
      a[i] += a[be];
    }
  }
  if(a[be] < 0) {
    for(int i = n; i >= 2; --i) {
      d[++res] = (data) {i, i - 1};
      a[i - 1] += a[i];
    }  
  } else {
    for(int i = 2; i <= n; ++i) {
      d[++res] = (data) {i - 1, i};
      a[i] += a[i - 1];
    }
  }
  printf("%d\n", res);
  for(int i = 1; i <= res; ++i) {
    printf("%d %d\n", d[i].x, d[i].y);
  }
  return 0;
}