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 <cstdio>
#include <algorithm>
#include <queue>
using namespace std;

int A[55];
vector<pair<int,int>> v;
int main(){
	int N; scanf("%d",&N);
	for(int i=1;i<=N;i++){
		scanf("%d",A+i);
	}
	int *mx = max_element(A+1,A+N+1);
	int *mn = min_element(A+1,A+N+1);
	if(*mx > 0 && *mn < 0){
		if(abs(*mx) > abs(*mn)){
			for(int i=1;i<=N;i++) v.push_back({mx-A,i});
			for(int i=1;i<N;i++) v.push_back({i,i+1});
		}
		else{
			for(int i=1;i<=N;i++) v.push_back({mn-A,i});
			for(int i=N;i>0;i--) v.push_back({i,i-1});
		}
	}
	else if(*mn >= 0){
		for(int i=1;i<N;i++) v.push_back({i,i+1});		
	}
	else {
		for(int i=N;i>0;i--) v.push_back({i,i-1});				
	}
	printf("%d\n",v.size());
	for(auto it : v){
		printf("%d %d\n",it.first,it.second);
	}
}