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Python 3 solution for cases where a+b is even (all test cases):

def solve(a, b, t):
    # Write your code here
    return int(pow(int((a+b)/2), t, 10**9 + 7))

static int solve(int a, int b, long t) {

    if(t == 1)return (int) ((a + b) / 2);

    return (int)pow((a+b)/2,t);
}
static long pow(long avg, long t) {
    long x = 1, n = avg;
    while(t > 0) {
        if(t%2 == 1) {
            x=(x * n)%1000000007;
        }
        n = (n * n)%1000000007;
        t /= 2;
    }
    return x;
}

`

GIven probability is 0.5 ( 1/2 ) . Factors are a and b . Initially cell is 1

Every time ,it grows according expected value using factor given a & b

fIrst time virus cell will be (a*(1/2)) + (b*(1/2)) = (a+b)/2;

Now cells are (a+b)/2; Now this will be growing into again by factor a & b with probability (1/2) .Let's see ....................

a*( ( a + b ) / 2 ) * ( 1 / 2 ) + b* ( ( a + b ) / 2 ) * ( 1 / 2 )

= ( ( a + b ) / 2 ) * (( a + b ) / 2 ) ) , when time is 2 milisecond .

So from this pattern , we can say ( ( a + b ) / 2 ) ^ t ....... %mod .

` My code

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ll;

const ll mod=1e9+7;

ll power(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%mod;

            a=(a*a)%mod;
            b=b>>1;
    }
    return ans;
}

int main()
{
    ll n,a,b,c,time;
    cin>>a>>b>>time;
    ll exp=(a+b)/2;
    ll result= power(exp,time)%mod;
    cout<<result<<endl;
}

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;
}