Let d(p,n,0) be the multiplicative inverse of n modulo prime p, defined as n × d(p,n,0) = 1 mod p.
Let d(p,n,k) = $\sum_{i=1}^n$d(p,i,k−1) for k ≥ 1.
Let D(a,b,k) = $\sum$(d(p,p-1,k) mod p) for all primes a ≤ p < a + b.

You are given:

• D(101,1,10) = 45
• D(10³,10²,10²) = 8334
• D(10⁶,10³,10³) = 38162302

Find D(10⁹,10⁵,10⁵).