Let $d(p,n,0)$ be the multiplicative inverse of $n$ modulo prime $p$, defined as $n\times d(p,n,0)=1modp$.
Let $d(p,n,k)=\sum _{i=1}^{n}d(p,i,k-1)$ for $k\ge 1$.
Let $D(a,b,k)=\sum (d(p,p-1,k)modp)$ for all primes $a\le p<a+b$.

You are given:

• $D(101,1,10)=45$
• $D({10}^3,{10}^2,{10}^2)=8334$
• $D({10}^6,{10}^3,{10}^3)=38162302$

Find $D({10}^9,{10}^5,{10}^5)$.