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<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
int n,a[100],sum;
int MIN,MAX,minflag,maxflag;
int main()
{
    n=sum=0;
    memset(a,0,sizeof(a));
    MIN=2000000000;
    MAX=-2000000000;
    scanf("%d",&n);
    for (int i=0; i<n; ++i)
    {
        scanf("%d",&a[i]);
        if (a[i]>MAX) {MAX=a[i];maxflag=i;}
        if (a[i]<MIN) {MIN=a[i];minflag=i;}
    }
    printf("%d\n",2*n);
    if (MAX>=0 && MAX >= -MIN)
    {
        cout<<maxflag+1<<' '<<1<<endl;
        cout<<maxflag+1<<' '<<1<<endl;
        for (int i=0; i<n-1; ++i)
        {
           cout<<i+1<<' '<<i+2<<endl;
           cout<<i+1<<' '<<i+2<<endl;
        }
    }else
    {
        cout<<minflag+1<<' '<<n<<endl;
        cout<<minflag+1<<' '<<n<<endl;
        for (int i=n; i>1; --i)
        {
            cout<<i<<' '<<i-1<<endl;
            cout<<i<<' '<<i-1<<endl;
        }
    }
    return 0;
}