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 DEBUG
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#define MAX_T 202
#define MAX_N 100

int N;
long long int a[1000000];

int main() {
#ifdef DEBUG

#endif
	scanf("%d", &N);
	int i;
	for (i = 0; i < N; i++)scanf("%lld", &a[i]);

	int max = 0;
	int tmp;
	int flag = 0;
	int idx;
	for (i = 0; i < N; i++) {
		tmp = abs((int)a[i]);
		if (max < tmp) {
			max = tmp;
			idx = i;
			if(a[i]>0) {
				flag = 1;
			}else{
				flag = -1;
			}
		}
	}
	int cnt=0;
	int ans[100][2];
	int ansIdx = 0;
	if (flag == 1) {
		for (i = 0; i < N; i++) {
			if (a[i] < 0) {
				a[i] += a[idx];
				ans[ansIdx][0] = i + 1;
				ans[ansIdx++][1] = idx + 1;
				cnt++;
			}
		}
		for (i = 0; i < N-1; i++) {
			if (a[i] > a[i + 1]) {
				a[i + 1] += a[i];
				ans[ansIdx][0] = i + 2;
				ans[ansIdx++][1] = i + 1;
				cnt++;
			}
		}
	}
	else{
		for (i = 0; i < N; i++) {
			if (a[i] > 0) {
				a[i] += a[idx];
				ans[ansIdx][0] = i + 1;
				ans[ansIdx++][1] = idx + 1;
				cnt++;
			}
		}
		for (i = N-1; i > 0; i--) {
			if (a[i] < a[i - 1]) {
				a[i - 1] += a[i];
				ans[ansIdx][0] = i;
				ans[ansIdx++][1] = i + 1;
				cnt++;
			}
		}
	}

	printf("%d\n", cnt);
	for (i = 0; i < cnt; i++) {
		printf("%ld %ld\n", ans[i][0], ans[i][1]);
	}
}