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<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e6+10;

int main()
{
	int n,k,s[55],a[110][2];
	while(~scanf("%d",&n))
	{
		for(int i=1;i<=n;i++) scanf("%d",&s[i]);
		int cnt=0;
		
		for(int i=1;i<n;i++)
		{
			if(s[i]>s[i+1])
			{
				if(s[i]==0)
				{
					a[cnt][0]=i+1;
					a[cnt][1]=i;
					cnt++;
				}
				else if(s[i+1]==0)
				{
					a[cnt][0]=i;
					a[cnt][1]=i+1;
					cnt++;
				}
				else if(s[i]>0&&s[i+1]>0)
				{
					a[cnt][0]=i;
					a[cnt][1]=i+1;
					cnt++;
				}
				else if(s[i]<0&&s[i+1]<0)
				{
					a[cnt][0]=i+1;
					a[cnt][1]=i;
					cnt++;
				}
				else if(s[i]>0&&s[i+1]<0)
				{
					if(abs(s[i])>=abs(s[i+1]))
					{
						a[cnt][0]=i;
						a[cnt][1]=i+1;
						cnt++;
						a[cnt][0]=i;
						a[cnt][1]=i+1;
						cnt++;
					}
					else
					{
						a[cnt][0]=i+1;
						a[cnt][1]=i;
						cnt++;
						a[cnt][0]=i+1;
						a[cnt][1]=i;
						cnt++;
					}
				}
			}
		}
		printf("%d\n",cnt);
		for(int i=0;i<cnt;i++)
		printf("%d %d\n",a[i][0],a[i][1]);
	}
	return 0;
}