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 <iostream>
#include <stdio.h>
#include <string.h>
#include <vector>
#define MAX_N 55
#define INF 1000000000

using namespace std;

struct Data
{
	int x,y;
	Data(int _x,int _y) { x=_x,y=_y; }
	Data(){}
};

int n;
int mx=-INF,mxid;
int mn=INF,mnid;
int a[MAX_N];
vector<Data> v;

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		if(a[i]>mx) mx=a[i],mxid=i;
		if(a[i]<mn) mn=a[i],mnid=i;
	}
	if(mn>=0 || (mx>=0 && mx>=(-mn)))
	{
		for(int i=1;i<=n;i++)
		{
			if(a[i]<0)
			{
				a[i]+=mx;
				v.push_back(Data(mxid,i));
			}
		}
		for(int i=2;i<=n;i++)
		{
			a[i]+=a[i-1];
			v.push_back(Data(i-1,i));
		}
	}
	else
	{
		for(int i=1;i<=n;i++)
		{
			if(a[i]>0)
			{
				a[i]+=mn;
				v.push_back(Data(mnid,i));
			}
		}
		for(int i=n-1;i>=1;i--)
		{
			a[i]+=a[i+1];
			v.push_back(Data(i+1,i));
		}
	}
	printf("%d\n",(int)v.size());
	for(int i=0;i<v.size();i++) printf("%d %d\n",v[i].x,v[i].y);
	/*for(int i=1;i<=n;i++) cout<<a[i]<<" "; cout<<endl;
	for(int i=2;i<=n;i++) if(a[i]<a[i-1]) { cout<<"WA"<<endl; return 0; }
	cout<<"AC"<<endl;*/
}