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>

using namespace std;
using ll = long long;
constexpr ll MOD = 1000000007;

/* SOLVE NOTES
(S + a) / n >= (T - b) / m
then
S / n < T / m

we have 
S / n - T / m >= - a / n - b / m
S / n - T / m < 0 

  3 -3      
1 5 -5 2 => 1 0 0 2
            1/2 1/2 0 2
            1/2 1/4 1/4 2
            1/3 1/3 1/3 2

A x y
P: A A+x A+x+y

[x > y]
A x+y/2 x+y/2
P: A A+(x+y)/2 A+x+y

YOU TAKE THE LOWER HULL!
*/

int main() {
    int n; scanf("%d", &n);
    vector<int> c(n), b(n-1);
    for (int i = 0; i < n; ++i) scanf("%d", &c[i]);
    for (int i = 0; i < n - 1; ++i) scanf("%d", &b[i]);
    vector<int> bDeduction(n);
    for (int i = 0; i < n - 1; ++i) bDeduction[i + 1] = bDeduction[i] + b[i];
    for (int i = 0; i < n - 1; ++i) bDeduction[i + 1] += bDeduction[i];
    int q; scanf("%d", &q);
    ll maxSum = accumulate(begin(c), end(c), 0LL);
    while (q--) {
        int x; scanf("%d", &x);
        // count the number of ways to be at i, with a partial sum of P,
        // without going below the (0, 0) - (1, x) line

        // consider a[0], a[1] - b[0], a[2] - b[0] - b[1], ...
        // so we only have to average points
        // then consider the partial sums of these
        // so 0, a[0], a[0] + a[1] - b[0], ...
        // so an average becomes a take-the-midpoint
        // but this ONLY works if you're bumping down a concave section
        // maxsum here is without bDeduction

        // x y z
        // 0, x, x + y - 2, x + y + z - 5
        // 0 1 4
        // 0 -1 1
        // 0 0 -1 0
        // 0 -1/2 -1 0
        // 0 -1/2 -1/2 1
        vector<vector<ll>> waysToBeAt(n + 1, vector<ll>(maxSum + 1));
        waysToBeAt[0][0] = 1;
        for (int elem = 1; elem <= n; ++elem) {
            for (int s = 0; s <= maxSum; ++s) {
                int actualYC = s - bDeduction[elem-1];
                if (actualYC < x * elem) continue;
                for (int vTaken = 0; vTaken <= c[elem-1]; ++vTaken) {
                    if (s - vTaken < 0) continue;
                    waysToBeAt[elem][s] = (waysToBeAt[elem][s] + 
                            waysToBeAt[elem-1][s-vTaken]) % MOD;
                }
            }
        }
        printf("%lld\n", accumulate(begin(waysToBeAt[n]), end(waysToBeAt[n]), 0LL) % MOD);
    }
}



