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>
#define MAIN 110
#define INF 0X3F3F3F3F
using namespace std;
int main()
{
    int n,a[110];
    cin>>n;
    int maxx=-INF,minn=INF;
    int maxnum,minnum;
    for(int i=1; i<=n; i++)
    {
        cin>>a[i];
        if(maxx<a[i])
        {
            maxx=a[i];
            maxnum=i;
        }
        if(minn>a[i])
        {
            minn=a[i];
            minnum=i;
        }
    }
    if(minn>=0)
    {
        cout<<n-1<<endl;
        for(int i=1; i<n; i++)
            cout<<i<<" "<<i+1<<endl;
        return 0;
    }
    if(abs(minn)<=abs(maxx))
    {
        cout<<n*2-1<<endl;
        for(int i=1; i<=n; i++)
            cout<<maxnum<<" "<<i<<endl;
        for(int i=1; i<n; i++)
            cout<<i<<" "<<i+1<<endl;
        return 0;
    }
    cout<<n*2-1<<endl;
    for(int i=1; i<=n; i++)
        cout<<minnum<<" "<<i<<endl;
    for(int i=n; i>1; i--)
        cout<<i<<" "<<i-1<<endl;
    return 0;
}
