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 (Clang 3.8.0) (judge verdict: Accepted)

#include "stdio.h"
#include "math.h"

int N;
int a[50];

 
int main(void){
	int i;
	int max = 0xFFFFFFFF;
	int min = 0x7FFFFFFF;
	int max_ind = 0;
	int min_ind = 0;
	int to_plus = 0;

	scanf("%d", &N);

	for(i=0;i<N;i++){
		scanf("%d", &a[i]);
		if(a[i]<min) min = a[i], min_ind = i;
		if(a[i]>max) max = a[i], max_ind = i;
	}

	if(abs(min)<abs(max)){
		to_plus = 1;
	}

printf("%d\n", N + N -1);

	if(to_plus){
		for(i=0;i<N;i++){
printf("%d %d\n", max_ind+1, i+1);
//			a[i]+=max;
		}
	}
	else //to minus
	{
		for(i=N-1;i>=0;i--){
printf("%d %d\n", min_ind+1, i+1);
//			a[i]+=min;
		}
	}


	if(to_plus){
		for(i=1;i<N;i++){
printf("%d %d\n", i, i+1);
//			a[i]+=a[i-1];
		}
	}
	else //to minus
	{
		for(i=N-1;i>0;i--){
printf("%d %d\n", i+1, i);
//			a[i]+=a[i+1];
		}
	}


//	printf("%d\n", sum);
	return 0;
}