This is the easy version of the problem. The only difference is that in this version q = 1. You can make hacks only if both versions of the problem are solved.

There is a process that takes place on arrays a and b of length n and length n-1 respectively. 

The process is an infinite sequence of operations. Each operation is as follows: 

  * First, choose a random integer i (1 ≤ i ≤ n-1). 
  * Then, simultaneously set a_i = min\left(a_i, \frac{a_i+a_{i+1}-b_i}{2}\right) and a_{i+1} = max\left(a_{i+1}, \frac{a_i+a_{i+1}+b_i}{2}\right) without any rounding (so values may become non-integer). 

See notes for an example of an operation.

It can be proven that array a converges, i. e. for each i there exists a limit a_i converges to. Let function F(a, b) return the value a_1 converges to after a process on a and b.

You are given array b, but not array a. However, you are given a third array c. Array a is good if it contains only integers and satisfies 0 ≤ a_i ≤ c_i for 1 ≤ i ≤ n.

Your task is to count the number of good arrays a where F(a, b) ≥ x for q values of x. Since the number of arrays can be very large, print it modulo 10^9+7.

Input

The first line contains a single integer n (2 ≤ n ≤ 100).

The second line contains n integers c_1, c_2 …, c_n (0 ≤ c_i ≤ 100).

The third line contains n-1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 100).

The fourth line contains a single integer q (q=1).

The fifth line contains q space separated integers x_1, x_2, …, x_q (-10^5 ≤ x_i ≤ 10^5).

Output

Output q integers, where the i-th integer is the answer to the i-th query, i. e. the number of good arrays a where F(a, b) ≥ x_i modulo 10^9+7.

Example

Input


3
2 3 4
2 1
1
-1


Output


56

Note

The following explanation assumes b = [2, 1] and c=[2, 3, 4] (as in the sample).

Examples of arrays a that are not good: 

  * a = [3, 2, 3] is not good because a_1 > c_1; 
  * a = [0, -1, 3] is not good because a_2 < 0. 



One possible good array a is [0, 2, 4]. We can show that no operation has any effect on this array, so F(a, b) = a_1 = 0.

Another possible good array a is [0, 1, 4]. In a single operation with i = 1, we set a_1 = min((0+1-2)/(2), 0) and a_2 = max((0+1+2)/(2), 1). So, after a single operation with i = 1, a becomes equal to [-1/2, 3/2, 4]. We can show that no operation has any effect on this array, so F(a, b) = -1/2.
#include <bits/stdc++.h>

#define vi vector<int>
#define vvi vector<vi>
#define ll long long
#define vll vector<ll>
#define vvll vector<vll>
#define ii pair<int, int>
#define vii vector<ii>
#define vvii vector<vii>
#define fi first
#define se second
#define all(v) v.begin(), v.end()
#define uni(v) v.erase(unique(v.begin(), v.end()), v.end())
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) (ll) a / __gcd(a, b) * b
#define prt(u, v) \
    for (auto x = u; x != v;) cout << (*x) << " \n"[++x == v];

using namespace std;

void print();
template <typename T, typename... Args>
void print(T x, Args... args);

typedef vector<double> vd;

void apply(vd &a, vd &b, int p) {
    double u = a[p];
    double v = a[p + 1];
    a[p] = min(u, (u + v - b[p]) / 2);
    a[p + 1] = max(v, (u + v + b[p]) / 2);
}
void test() {
    vd a = {100, 2, 30};
    vd b = {0, 0};

    prt(a.begin(), a.end());
    prt(b.begin(), b.end());

    srand(time(NULL));
    for (int i = 0; i < 10000; ++i) apply(a, b, rand() % 2);

    prt(a.begin(), a.end());
}
void solve() {
    const int M = 1e9 + 7;
    int n;
    cin >> n;
    vi c(n), b(n - 1);
    for (int i = 0; i < n; ++i) cin >> c[i];
    for (int i = 0; i < n - 1; ++i) cin >> b[i];

    int q;
    cin >> q;
    vi x(q);
    for (int i = 0; i < q; ++i) cin >> x[i];

    vi a(n);
    a[0] = x[0];
    for (int i = 0; i < n - 1; ++i) a[i + 1] = a[i] + b[i];
    partial_sum(a.begin(), a.end(), a.begin());

    const int N = 107;
    const int V = N * N;

    vi dp(V), tmp(V);
    dp[0] = 1;
    for (int i = 0; i < n; ++i) {
        copy(dp.begin(), dp.end(), tmp.begin());
        fill(dp.begin(), dp.end(), 0);

        for (int j = 0; j < V; ++j)
            if (tmp[j]) {
                int low = max(a[i], j);
                int hig = max(a[i], j + c[i] + 1);

                if (low < V) dp[low] = (dp[low] + tmp[j]) % M;
                if (hig < V) dp[hig] = (dp[hig] + M - tmp[j]) % M;
            }

        partial_sum(dp.begin(), dp.end(), dp.begin(),
                    [](int u, int v) { return (u + v) % M; });
    }

    print(accumulate(dp.begin(), dp.end(), 0LL,
                     [](int u, int v) { return (u + v) % M; }));
}
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

    int t = 1;
    // cin >> t;
    for (int i = 0; i++ < t;) solve();
    // test();

    return 0;
}

void print() { cout << "\n"; }
template <typename T, typename... Args>
void print(T x, Args... args) {
    if (sizeof...(args)) {
        cout << x << ' ';
        print(args...);
    } else {
        cout << x << '\n';
    }
}