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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <math.h>
#include <limits.h>
#include <assert.h>
#define FOR(i,a,n) for(i=(a);i<(n);i++)
#define swap(type,a,b) do{type t=a;a=b;b=t;}while(0)
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))
#define ll long long
#define INF 100000000
#define MOD 1000000007
#define SIZE 100005
#define LOCAL
#ifdef LOCAL
    #define DEBUG(...) printf(" "__VA_ARGS__)
#else
    #define DEBUG(...)
#endif /* LOCAL */
int comp(const void* a,const void* b){
	return *(int*)a-*(int*)b;
}
int a[50];
int res[100][2];
int main(void)
{
	int n,min=INF,max=0,min_id,max_id,s=0,i;
	scanf("%d",&n);
	FOR(i,0,n){
		scanf("%d",a+i);
		if(min>a[i]) min=a[i],min_id=i;
		if(max<a[i]) max=a[i],max_id=i;
	}
	if(-min>max){
		for(i=n-2;i>=0;i--){
			if(a[i]>a[i+1]){
				while(a[i]>a[i+1]){
					res[s][0]=min_id+1; res[s++][1]=i+1;
					a[i]+=min;
					if(a[i]<min) min=a[i],min_id=i;
				}
			}
		}
	}
	else{
		FOR(i,1,n){
			if(a[i-1]>a[i]){
				while(a[i-1]>a[i]){
					res[s][0]=max_id+1; res[s++][1]=i+1;
					a[i]+=max;
					if(a[i]>max) max=a[i],max_id=i;
				}
			}
		}
	}
	printf("%d\n",s);
	FOR(i,0,s){
		printf("%d %d\n",res[i][0],res[i][1]);
	}
	return 0;
}