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;
int num,d[120][3];
int main()
{
	int n;
	cin>>n;
	int a[n+1];
	int i,maxv=-0x7fffffff,p;
	bool flag=true;
	for(i=1;i<=n;i++)
	{
		cin>>a[i];
		if(abs(a[i])>maxv)
		{
			maxv=abs(a[i]);
			p=i;
			if(a[i]<0)
			{
				flag=false;
			}
			else
			{
				flag=true;
			}
		}
	}
	for(i=1;i<=n;i++)
	{
		num++;
		d[num][1]=p;
		d[num][2]=i;
		a[i]+=a[p];
	}
	if(flag)
	{
		for(i=2;i<=n;i++)
		{
			num++;
			d[num][1]=i-1;
			d[num][2]=i;
		}
	}
	else
	{
		for(i=n;i>=2;i--)
		{
			num++;
			d[num][1]=i;
			d[num][2]=i-1;
		}
	}
	cout<<num<<endl;//2*n-1;
	for(i=1;i<=num;i++)
	{
		cout<<d[i][1]<<' '<<d[i][2]<<endl;
	}
	return 0;
}