Let us consider mixtures of three substances: A, B and C. A mixture can be described by a ratio of the amounts of A, B, and C in it, i.e., $(a:b:c)$. For example, a mixture described by the ratio $(2:3:5)$ contains $20\mathrm{\%}$ A, $30\mathrm{\%}$ B and $50\mathrm{\%}$ C.

For the purposes of this problem, we cannot separate the individual components from a mixture. However, we can combine different amounts of different mixtures to form mixtures with new ratios.

For example, say we have three mixtures with ratios $(3:0:2)$, $(3:6:11)$ and $(3:3:4)$. By mixing $10$ units of the first, $20$ units of the second and $30$ units of the third, we get a new mixture with ratio $(6:5:9)$, since:
$(10\cdot \frac{3}{5}$ + $20\cdot \frac{3}{20}+30\cdot \frac{3}{10}:10\cdot \frac{0}{5}+20\cdot \frac{6}{20}+30\cdot \frac{3}{10}:10\cdot \frac{2}{5}+20\cdot \frac{11}{20}+30\cdot \frac{4}{10})=(18:15:27)=(6:5:9)$

However, with the same three mixtures, it is impossible to form the ratio $(3:2:1)$, since the amount of B is always less than the amount of C.

Let $n$ be a positive integer. Suppose that for every triple of integers $(a,b,c)$ with $0\le a,b,c\le n$ and $gcd(a,b,c)=1$, we have a mixture with ratio $(a:b:c)$. Let $M(n)$ be the set of all such mixtures.

For example, $M(2)$ contains the $19$ mixtures with the following ratios:

$$\begin{array}{rl}\{& (0:0:1),(0:1:0),(0:1:1),(0:1:2),(0:2:1),\\ & (1:0:0),(1:0:1),(1:0:2),(1:1:0),(1:1:1),\\ & (1:1:2),(1:2:0),(1:2:1),(1:2:2),(2:0:1),\\ & (2:1:0),(2:1:1),(2:1:2),(2:2:1)\}.\end{array}$$

Let $E(n)$ be the number of subsets of $M(n)$ which can produce the mixture with ratio $(1:1:1)$, i.e., the mixture with equal parts A, B and C.
We can verify that $E(1)=103$, $E(2)=520447$, $E(10)mod{11}^8=82608406$ and $E(500)mod{11}^8=13801403$.
Find $E(10000000)mod{11}^8$.