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.

Examples

Input

3
-2 5 -1


Output

2
2 3
3 3


Input

2
-1 -3


Output

1
2 1


Input

5
0 0 0 0 0


Output

0
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
#define fi first
#define se second
#define rep(i,n) for(int i = 1;i < n;i++)
#define drep(i,n) for(int i = n;i > 1;i--)
typedef pair<int,int> P;

int a[51];
pair<int,int> mn, mx;
int main(){
	int n, t, f;
	mn.fi = inf, mx.fi = -inf;
	scanf("%d", &n);
	for(int i = 0;i < n;i++){
		scanf("%d", &a[i]);
		if(mn.fi > a[i])mn = P(a[i], i+1);
		if(mx.fi < a[i])mx = P(a[i], i+1);
	}
	
	printf("%d\n", 2*n-1);
	if(abs(mn.fi) <= abs(mx.fi))t = mx.se, f = 0;
	else t = mn.se, f = 1;
	
	for(int i = 0;i < n;i++){
		printf("%d %d\n", t, i+1);
	}
	
	if(f == 0){
		rep(i,n)printf("%d %d\n", i, i+1);
	}else{
		drep(i,n)printf("%d %d\n", i, i-1);
	}
	return 0;
}