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

This method doesn't use recursion with O(log(n)) complexity. Check this out.

#define ull unsigned long long
#define MODULO 1000000007

ull PowMod(ull n)
{
    ull ret = 1;
    ull a = 2;
    while (n > 0) {
        if (n & 1) ret = ret * a % MODULO;
        a = a * a % MODULO;
        n >>= 1;
    }
    return ret;
}

And this is pseudo from Wikipedia (see Right-to-left binary method section)

function modular_pow(base, exponent, modulus)
Assert :: (modulus - 1) * (base mod modulus) does not overflow base
result := 1
base := base mod modulus
while exponent > 0
    if (exponent mod 2 == 1):
       result := (result * base) mod modulus
    exponent := exponent >> 1
    base := (base * base) mod modulus
return result