Problem: 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.

Solution (in JAVA):

Solution:


import java.util.Scanner;

public class c1540easy {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);

        solve(s.nextInt(), s);
    }


    static long mod = 1000000007;
    public static void solve(int l, Scanner s) {
        int c[] = new int[l];
        for (int i = 0; i < l; i++) {
            c[i] = s.nextInt();
        }

        int b[] = new int[l - 1];
        for (int i = 0; i < l - 1; i++) {
            b[i] = s.nextInt();
        }

        s.nextInt();
        int q = s.nextInt();

        long next[] = new long[]{1};
        long last[];

        int min = q;

        int sum;

        int bsum = 0;

        for (int i = 0; i < l; i++) {
            last = next;
            next = new long[last.length + c[i] + 1];

            for (int j = 0; j < last.length; j++) {
                for (int k = 0; k <= c[i]; k++) {
                    sum = k + j;
                    if (sum >= min) {
                        next[sum] = (next[sum] + last[j]) % mod;
                    }
                }
            }

            bsum += i != l - 1 ? b[i] : 0;
            min += q + bsum;
        }

        long erg = 0;
        for (int i = 0; i < next.length; i++) {
            erg = (erg + next[i]) % mod;
        }

        System.out.println(erg);
    }
}