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: Accepted)

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
using namespace std;

int n;
const int maxn = 55;
int a[maxn];

int main(){
	cin >> n;
	int id = 0,ans = 0;
	for(int i = 1;i <= n;i++){
		scanf("%d",&a[i]);
		if(abs(a[i]) > ans){
			ans = abs(a[i]);
			id = i;
		}
	}
	if(a[id] == 0){
		puts("0");
	}else if(a[id] > 0){
		printf("%d\n",2 * (n - 1));
		for(int i = 2;i <= n;i++){
			printf("%d %d\n",id,i);	
			printf("%d %d\n",id,i);	
			id = i;
		}	
	}else{
		printf("%d\n",2 * (n - 1));
		for(int i = n - 1;i >= 1;i--){
			printf("%d %d\n",id,i);	
			printf("%d %d\n",id,i);	
			id = i;
		}	
	}
	return 0;
}