Initially there is a single virus who replicates into (a+b)/2 expected number of viruses at time=1 ms. These (a+b)/2 new viruses further replicate into (a+b)/2 new viruses each at time=2 ms.
Let us call (a+b)/2 as mean.
Total number of viruses at time=2ms:
=mean*mean
=mean^2
After time=t ms, mean^t;
A simple pow(mean,t) should work if value of t is small. However, in this case, we need to use fast exponentiation.

long calc(int n, long long pow)      //fast exponentiation 
{
    if(pow==0)
        return 1;
    if(pow==1)
        return n;
    long p=calc(n,pow/2);
    p=(p*p)%mod;
    if(pow%2!=0)
        p=(p*n)%mod;
    return p;

}
int main() {
   long long t;
    int a,b;
    cin>>a>>b>>t;
    int mean=(a+b)/2;
    long p=calc(mean,t);
    cout<<p;
    return 0;
}