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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main()
{
	int n;
	scanf("%d", &n);
	int i;
	long long int a[55];
	for (i = 0; i < n; i++)
		scanf("%lld", &a[i]);
	long long int max, min;
	max = min = 0;
	for (i = 0; i < n; i++)
	{
		if (a[i] > a[max])
			max = i;
		if (a[i] < a[min])
			min = i;
	}
	int m, ans[102][2];
	m = 0;
	if (a[max] + a[min] >= 0)
	{
		for (i = 0; i < n; i++)
		{
			if (a[i] < 0)
			{
				ans[m][0] = max;
				ans[m][1] = i;
				m++;
				a[i] += a[max];
			}
		}
		for (i = 0; i < n - 1; i++)
		{
			if (a[i] > a[i + 1])
			{
				ans[m][0] = i;
				ans[m][1] = i + 1;
				m++;
				a[i + 1] += a[i];
			}
		}
	}
	else
	{
		for (i = 0; i < n; i++)
		{
			if (a[i] >= 0)
			{
				ans[m][0] = min;
				ans[0][1] = i;
				m++;
				a[i] += a[min];
			}
		}
		for (i = n - 1; i > 0; i--)
		{
			if (a[i - 1] > a[i])
			{
				ans[m][0] = i;
				ans[m][1] = i - 1;
				m++;
				a[i - 1] += a[i];
			}
		}
	}
	printf("%d\n", m);
	for (i = 0; i < m; i++)
		printf("%d %d\n", ans[i][0] + 1, ans[i][1] + 1);
	return 0;
}