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>
#include <stdlib.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define INF ((1LL<<62)-(1LL<<31))

ll a[99],x[99],y[99];
ll ans;
int main(){
	int n;
	scanf("%d",&n);
	rep(i,0,n)scanf("%lld",a+i);
	
	ll M=-INF,m=INF,Mi,mi;
	rep(i,0,n){
		if(a[i]>M){
			M=a[i];
			Mi=i;
		}
		if(a[i]<m){
			m=a[i];
			mi=i;
		}
	}
	if(M>0&&m<0){
		if(M>=-m)rep(i,0,n){x[ans]=Mi;y[ans]=i;ans++;m+=M;}//全部0以上にする
		else rep(i,0,n){x[ans]=mi;y[ans]=i;ans++;M+=m;}//全部0以下にする
	}
	
	//累積和をとる
	if(m>=0)rep(i,1,n){x[ans]=i-1;y[ans]=i;ans++;}//全部0以上のとき
	else for(int i=n-1;i>=1;i--){x[ans]=i;y[ans]=i-1;ans++;}//全部0以下のとき
	
	printf("%lld\n",ans);
	rep(i,0,ans)printf("%lld %lld\n",x[i]+1,y[i]+1);
	return 0;
}