/*
Matrix Exponentiation.
The problem can be solved with DP but constraints are high.
ai = bi (for i <= k)
ai = c1*ai-1 + c2*ai-2 + ... + ck*ai-k (for i > k)
Taking the example of Fibonacci series, K=2
b1 = 1, b2=1
c1 = 1, c2=1
a = 0 1 1 2 ....
This way you can find the 10^18 fibonacci number%MOD.
I have given a general way to use it. The program takes the input of B and C
matrix.
Steps for Matrix Expo
1. Create vector F1 : which is the copy of B.
2. Create transpose matrix (Learn more about it on the internet)
3. Perform T^(n-1) [transpose matrix to the power n-1]
4. Multiply with F to get the last matrix of size (1xk).
The first element of this matrix is the required result.
*/

#include <iostream>
#include <vector>

using std::cin;
using std::cout;
using std::vector;

#define ll int64_t
#define endl '\n'
#define pb push_back
#define MOD 1000000007
ll ab(ll x) { return x > 0LL ? x : -x; }
ll k;
vector<ll> a, b, c;

// To multiply 2 matrix
vector<vector<ll>> multiply(vector<vector<ll>> A, vector<vector<ll>> B) {
    vector<vector<ll>> C(k + 1, vector<ll>(k + 1));
    for (ll i = 1; i <= k; i++) {
        for (ll j = 1; j <= k; j++) {
            for (ll z = 1; z <= k; z++) {
                C[i][j] = (C[i][j] + (A[i][z] * B[z][j]) % MOD) % MOD;
            }
        }
    }
    return C;
}

// computing power of a matrix
vector<vector<ll>> power(vector<vector<ll>> A, ll p) {
    if (p == 1) return A;
    if (p % 2 == 1) {
        return multiply(A, power(A, p - 1));
    } else {
        vector<vector<ll>> X = power(A, p / 2);
        return multiply(X, X);
    }
}

// main function
ll ans(ll n) {
    if (n == 0) return 0;
    if (n <= k) return b[n - 1];
    // F1
    vector<ll> F1(k + 1);
    for (ll i = 1; i <= k; i++) F1[i] = b[i - 1];

    // Transpose matrix
    vector<vector<ll>> T(k + 1, vector<ll>(k + 1));
    for (ll i = 1; i <= k; i++) {
        for (ll j = 1; j <= k; j++) {
            if (i < k) {
                if (j == i + 1)
                    T[i][j] = 1;
                else
                    T[i][j] = 0;
                continue;
            }
            T[i][j] = c[k - j];
        }
    }
    // T^n-1
    T = power(T, n - 1);

    // T*F1
    ll res = 0;
    for (ll i = 1; i <= k; i++) {
        res = (res + (T[1][i] * F1[i]) % MOD) % MOD;
    }
    return res;
}

// 1 1 2 3 5

int main() {
    cin.tie(0);
    cout.tie(0);
    ll t;
    cin >> t;
    ll i, j, x;
    while (t--) {
        cin >> k;
        for (i = 0; i < k; i++) {
            cin >> x;
            b.pb(x);
        }
        for (i = 0; i < k; i++) {
            cin >> x;
            c.pb(x);
        }
        cin >> x;
        cout << ans(x) << endl;
        b.clear();
        c.clear();
    }
    return 0;
}