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<bits/stdc++.h>
#define int long long
using namespace std;
int N,p[50];
signed main(){
	cin>>N;
	for(int i=0;i<N;i++)
		cin>>p[i];
	bool f;
	int u=0,v;
	for(int i=0;i<N;i++)
		if(u<abs(p[i])){
			u=abs(p[i]);
			v=i;
			f=(p[i]<0);
		}
	if(u==0){
		cout<<0<<endl;
		return 0;
	}
	if(f){
		for(int i=0;i<N;i++)
			p[i]=-p[i];
		reverse(p,p+N);
		v=N-v-1;
	}
	vector<pair<int,int>> ve;
	for(int i=1;i<N;i++)
		if(p[i-1]>p[i]){
			p[i]+=u;
			ve.push_back({v,i});
			if(p[i-1]>p[i]){
				p[i]+=u;
				ve.push_back({v,i});
			}
			u=p[i];
			v=i;
		}
	cout<<ve.size()<<endl;
	for(auto x:ve)
		if(f)
			cout<<N-x.first<<" "<<N-x.second<<endl;
		else
			cout<<x.first+1<<" "<<x.second+1<<endl;
}