1. Abridged Problem Statement  
Given K test cases, each with an integer N (1 ≤ N ≤ 1000) followed by a sequence A₁…A_N (0 ≤ Aᵢ ≤ 10⁹). Define  
S = A₁ + A₁·A₂ + A₁·A₂·A₃ + … + A₁·A₂·…·A_N.  
Compute the digital root of S, where the digital root is obtained by repeatedly summing the decimal digits until a single digit remains (0’s digital root is 0).

2. Detailed Editorial  

Overview of Digital Root and Modulo 9  
- The digital root dr(x) of a positive integer x is x mod 9, except that multiples of 9 map to 9. Formally, for x>0:  
  dr(x) = 1 + (x−1) mod 9; dr(0) = 0.  
- Equivalently, dr(x) =  x mod 9, then if result=0 and x>0 output 9; if x=0 output 0.  

Problem Reduction  
We need dr(S) where  
  S = ∑_{i=1..N} P_i,   P_i = ∏_{j=1..i} A_j.  
Instead of computing huge products and sums, we work modulo 9 throughout:  

1. Handle the zero edge-case:  
   - If A₁=0, then every P_i = 0 and S=0 ⇒ dr(S)=0.  

2. Otherwise, maintain two variables:  
   - s = current prefix product modulo 9, initialized to 1.  
   - ans = running sum of s’s modulo 9, initialized to 0.  

3. Iterate i from 1 to N:  
   - Compute aᵢ_mod = Aᵢ mod 9.  
   - Update s = (s * aᵢ_mod) mod 9.  
   - Update ans = (ans + s) mod 9.  

4. Finally, if ans=0 output 9, else output ans.  

Complexity  
Each test runs in O(N) time and O(1) extra space (besides reading input). With N≤1000 and K≤5, it is extremely fast.

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Maximum possible N from constraints (not strictly needed here).
const int MAXN = (1 << 20);

int n;              // Number of elements in the current test case
int a[MAXN];        // Array to store the sequence A

// Read one test case: first n, then n integers into array a[]
void read() {
    cin >> n;
    for(int i = 0; i < n; i++) {
        cin >> a[i];
    }
}

// Solve one test case
void solve() {
    // Edge-case: if the very first element is zero,
    // every prefix product is zero, so S = 0 ⇒ digital root = 0
    if (a[0] == 0) {
        cout << 0 << '\n';
        return;
    }

    int ans = 0;  // Will accumulate S mod 9
    int s   = 1;  // Current prefix product mod 9

    // Loop over all elements
    for (int i = 0; i < n; i++) {
        // Reduce A[i] modulo 9 to keep numbers small
        a[i] %= 9;
        // Update prefix product: s = previous s * a[i] (mod 9)
        s = (s * a[i]) % 9;
        // Add this prefix product to the running sum (mod 9)
        ans = (ans + s) % 9;
    }

    // Convert ans mod 9 into digital root:
    // if ans==0 then sum S was a multiple of 9 but nonzero ⇒ digital root is 9
    // (we already handled the true-zero case above)
    if (ans == 0) 
        cout << 9 << '\n';
    else 
        cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int K;            // Number of test cases
    cin >> K;
    while (K--) {
        read();
        solve();
    }
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys

def digital_root_sum_of_prefix_products(A):
    """
    Given list A, compute S = A[0] + A[0]*A[1] + ... + product(A[0..N-1]).
    Return digital root of S.
    """
    # If the first element is zero, all prefix products are zero => S=0 => dr=0
    if A[0] == 0:
        return 0

    s = 1    # current prefix product modulo 9
    ans = 0  # running sum of prefix products modulo 9

    for x in A:
        x_mod = x % 9         # reduce to modulo 9
        s = (s * x_mod) % 9   # update prefix product mod 9
        ans = (ans + s) % 9   # accumulate into sum mod 9

    # If ans==0, S was nonzero multiple of 9 ⇒ digital root = 9
    return 9 if ans == 0 else ans

def main():
    data = sys.stdin.read().strip().split()
    it = iter(data)
    K = int(next(it))
    out = []
    for _ in range(K):
        n = int(next(it))
        A = [int(next(it)) for _ in range(n)]
        dr = digital_root_sum_of_prefix_products(A)
        out.append(str(dr))
    print('\n'.join(out))

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
- The digital root dr(x) equals x mod 9 (with 0↦0, and nonzero multiples of 9↦9).  
- We need dr(S) for S = ∑ prefix-products of A.  
- Maintain `s = ∏prefix mod 9` and `ans = ∑s mod 9`.  
- If A₁=0 output 0. Else compute in O(N) and output `ans==0 ? 9 : ans`.