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: Accepted)

#include <stdio.h>

int main(void) {
	int N;
	int a[64];
	int i;
	int min, max;
	int minIdx, maxIdx;
	int sousaCount = 0;
	int sousa[128][2];
	if (scanf("%d", &N) != 1) return 1;
	for (i = 0; i < N; i++) {
		if (scanf("%d", &a[i]) != 1) return 1;
	}
	min = max = a[0];
	minIdx = maxIdx = 0;
	for (i = 1; i < N; i++) {
		if (min > a[i]) { min = a[i]; minIdx = i; }
		if (max < a[i]) { max = a[i]; maxIdx = i; }
	}
	if (min == 0 && max == 0) {
		puts("0");
		return 0;
	}
	if (min < 0 && 0 < max) {
		if (-min > max) {
			for (i = 0; i < N; i++) {
				if (a[i] >= 0) {
					sousa[sousaCount][0] = minIdx;
					sousa[sousaCount][1] = i;
					sousaCount++;
					a[i] += a[minIdx];
				}
			}
			max = 0;
		} else {
			for (i = 0; i < N; i++) {
				if (a[i] <= 0) {
					sousa[sousaCount][0] = maxIdx;
					sousa[sousaCount][1] = i;
					sousaCount++;
					a[i] += a[maxIdx];
				}
			}
			min = 0;
		}
	}
	if (min < 0) {
		if (a[N - 1] == 0) {
			sousa[sousaCount][0] = minIdx;
			sousa[sousaCount][1] = N - 1;
			sousaCount++;
		}
		for (i = N - 2; i >= 0; i--) {
			sousa[sousaCount][0] = i + 1;
			sousa[sousaCount][1] = i;
			sousaCount++;
		}
	} else {
		if (a[0] == 0) {
			sousa[sousaCount][0] = maxIdx;
			sousa[sousaCount][1] = 0;
			sousaCount++;
		}
		for (i = 1; i < N; i++) {
			sousa[sousaCount][0] = i - 1;
			sousa[sousaCount][1] = i;
			sousaCount++;
		}
	}
	printf("%d\n", sousaCount);
	for (i = 0; i < sousaCount; i++) {
		printf("%d %d\n", sousa[i][0] + 1, sousa[i][1] + 1);
	}
	return 0;
}