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>
using namespace std;
#define MAXN 55
int v[MAXN];
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int N;
	cin >> N;
	int me = 1e9, ma = -1e9;
	int ime, ima;
	for(int i = 0; i < N; i++){
		cin >> v[i];
		if(me > v[i]) me = v[i], ime = i;
		if(ma < v[i]) ma = v[i], ima = i;
	}
	cout << N+N - 2 << "\n";
	if(abs(me) > abs(ma)){
		for(int i = 0; i < N; i++){
			if(i != ime) {
				cout << ime + 1 << " " << i+1 << "\n";
				v[i] += v[ime];
			}
		}
		for(int i = N-1; i >= 1; i--){
			cout << i+1 << " " << i << "\n";
		}
	}
	else{
		for(int i = 0; i < N; i++){
			if(i != ima) {
				cout << ima + 1 << " " << i+1 << "\n";
				v[i] += v[ima];
			}
		}	
		for(int i = 1; i < N; i++){
			cout << i << " " << i+1 << "\n";
		}
	}
	return 0;
}