What is the fastest way to compute large power of 2 modulo a number

Question

For 1 <= N <= 1000000000, I need to compute 2N mod 1000000007, and it must be really fast!
My current approach i

Answer 1

It can be solved in O((log n)^2). Try this approach:-

unsigned long long int fastspcexp(unsigned long long int n)
{
    if(n==0)
        return 1;
    if(n%2==0)
        return (((fastspcexp(n/2))*(fastspcexp(n/2)))%1000000007);
    else
        return ( ( ((fastspcexp(n/2)) * (fastspcexp(n/2)) * 2) %1000000007 ) ); 
}

This is a recursive approach and is pretty fast enough to meet the time requirements in most of the programming competitions.