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>
#include <stdlib.h>
int main(void){
	int n,maxi,mini;
	int cnt=0;
	int cnt2=0;
	int max=0;
	int min=0;
	int a[50]={0};
	int b[50]={0};
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
		if(a[i]<0){
			cnt++;
		}
		if(max<a[i]){
			max=a[i];
			maxi=i;
		}
		if(min>a[i]){
			min=a[i];
			mini=i;
		}
	}
	if(cnt==0){
		printf("%d\n",n-1);
		for(int i=1;i<n;i++){
			printf("%d %d",i,i+1);
		}
	}else if(cnt==n){
		printf("%d\n",n-1);
		for(int i=n;i>0;i--){
			printf("%d %d",i,i-1);
		}
	}else{
		printf("%d\n",n*2-2);
		if(abs(max)>=abs(min)){
			for(int i=1;i<n;i++){
				if(i!=maxi){
					printf("%d %d",maxi,i);
				}
			}
			for(int i=1;i<n;i++){
				printf("%d %d",i,i+1);
			}
		}else{
			for(int i=1;i<n;i++){
				if(i!=mini){
					printf("%d %d",mini,i);
				}
			}
			for(int i=n;i>0;i--){
				printf("%d %d",i,i-1);
			}
		}
	}
	return 0;
}