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: Accepted)

#include <stdio.h>
#include <stdlib.h>

int a[55];

int main(){

	int N;

	scanf("%d",&N);

	int i,j;

	int ma=0;
	int idx;

	for(i=1;i<=N;i++){
		scanf(" %d",&a[i]);
		if(ma<abs(a[i])){
			ma=abs(a[i]);
			idx=i;
		}
	}

	if(ma!=0){

		for(i=1;i<=N;i++){
			if(i!=idx){
				a[i]+=a[idx];
			}
		}
		int ans=N-1;

		if(a[idx]>0){
			for(i=2;i<=N;i++){
				a[i]+=a[i-1];
				ans++;
			}
			printf("%d\n",ans);
			for(i=1;i<=N;i++){
				if(i!=idx){
					printf("%d %d\n",idx,i);
				}
			}
			for(i=2;i<=N;i++){
				printf("%d %d\n",i-1,i);
			}
		}
		else{
			for(i=N;i>=2;i--){
				a[i-1]+=a[i];
				ans++;
			}
			printf("%d\n",ans);
			for(i=1;i<=N;i++){
				if(i!=idx){
					printf("%d %d\n",idx,i);
				}
			}
			for(i=N;i>=2;i--){
				printf("%d %d\n",i,i-1);
			}
		}
	}
	else{
		printf("0\n");
	}



	return 0;
}