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<bits/stdc++.h>
using namespace std;
int n,a[55];
int main(){
	int minid = 1, maxid = 1;
	scanf("%d",&n);
	for(int i=1; i<=n; i++){
		scanf("%d",&a[i]);
		if(a[i] > a[maxid])maxid = i;
		if(a[i] < a[minid])minid = i;
	}
	printf("%d\n",2*n-2);
	if(a[minid] + a[maxid] >= 0){
		for(int i=1; i<n; i++)printf("%d %d\n",maxid, i);
		for(int i=1; i<n; i++)printf("%d %d\n",i,i+1);
	}else{
		for(int i=1; i<n; i++)printf("%d %d\n",minid, i);
		for(int i=n; i>1; i--)printf("%d %d\n",i,i-1);
	}
	return 0;
}