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;
int n;
int arr[100];
int main()
{
    cin>>n;int mi=1e7,pmi,mx=-1e7,pmx;
    for(int i=1;i<=n;i++)cin>>arr[i],mi=min(mi,arr[i]),mx=max(mx,arr[i]);
    for(int i=1;i<=n;i++) {
        if(arr[i]==mi)pmi=i;
        if(arr[i]==mx)pmx=i;
    }
    if(mi>=0) {//都大于等于0
        cout<<n-1<<endl;
        for(int i=1;i<n;i++)cout<<i<<" "<<i+1<<endl;
    } else if(mx<=0) {//都小于等于
        cout<<n-1<<endl;
        for(int i=n;i>1;i--)cout<<i<<" "<<i-1<<endl;
    } else {//有正有负
        cout<<2*n-1<<endl;
        if(abs(mi)>abs(mx)) {//转化为第二种
            for(int i=1;i<=n;i++)cout<<pmi<<" "<<i<<endl;
            for(int i=n;i>1;i--)cout<<i<<" "<<i-1<<endl;
        } else {//转化为第一种
            for(int i=1;i<=n;i++)cout<<pmx<<" "<<i<<endl;
            for(int i=1;i<n;i++)cout<<i<<" "<<i+1<<endl;
        }
    }
}
