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<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<cstdio>
using namespace std;
int n;
int main(){
    cin>>n;
    int mn=1e6+5,mx=-1e6-5,pos1,pos2;
    for (int i=1;i<=n;i++){
        int x;cin>>x;
        if (mn>x){
            mn=x;
            pos1=i;
        }
        if (mx<x){
            mx=x;
            pos2=i;
        }
    }
    printf("%d\n",2*n-1);
    if (abs(mn)<abs(mx)){
        for (int i=1;i<=n;i++){
            printf("%d %d\n",pos2,i);
        }
        for (int i=2;i<=n;i++){
            printf("%d %d\n",i-1,i);
        }
    }
    else{
        for (int i=1;i<=n;i++){
            printf("%d %d\n",pos1,i);
        }
        for (int i=n;i>=2;i--){
            printf("%d %d\n",i,i-1);
        }
    }
    return 0;
}
