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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
//#include <limits.h>

int main(void)
{
	int n;
	int i, max_i, min_i;
	int a[50];
//	int max = INT_MIN, min = INT_MAX;
	int max = -10000000, min = 10000000;
	int cs;
	int cnt;
	scanf("%d", &n);
	for (i = 0; i < n; ++i){
		scanf("%d", a + i);
		if (a[i] > max){
			max = a[i];
			max_i = i;
		}
		if (a[i] < min){
			min = a[i];
			min_i = i;
		}
	}
	if (max <= 0){
		//all negative
		printf("%d\n", n - 1);
		for (i = 0; i < n - 1; ++i){
			printf("%d %d\n", n - i, n - i - 1);
		}
	}
	else if (min >= 0){
		//all positive
		printf("%d\n", n - 1);
		for (i = 0; i < n - 1; ++i){
			printf("%d %d\n", i + 1, i + 2);
		}
	}
	else if (-max < min){
		printf("%d\n", n + n - 1);
		for (i = 0; i < n; ++i){
			printf("%d %d\n", i + 1, max_i);
		}
		for (i = 0; i < n - 1; ++i){
			printf("%d %d\n", i + 1, i + 2);
		}
	}
	else{
		printf("%d\n", n + n - 1);
		for (i = 0; i < n; ++i){
			printf("%d %d\n", i + 1, min_i);
		}
		for (i = 0; i < n - 1; ++i){
			printf("%d %d\n", n - i, n - i - 1);
		}
	}
	return 0;
}