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<cmath>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<map>
using namespace std;

int main() {
    int n, M=1000000007, tmp, maxc=100;
    vector<int> c;
    scanf("%d", &n);

    int sumc=0;
    for(int i=0;i<n;++i) {
        scanf("%d", &tmp);
        c.push_back(tmp);
        sumc+=tmp;
    }

    int sumb=0, global_bi_over_i_ub=0;
    vector<int> b(n, 0);
    for(int i=0;i<n-1;++i) {
        scanf("%d", &tmp);
        sumb = sumb+tmp;
        b[i+1] = b[i] + sumb;
        global_bi_over_i_ub = min(global_bi_over_i_ub, -(b[i+1] / (i+2)));
    }
    int global_x_ub = maxc+global_bi_over_i_ub;

    int q;
    vector<int> x;
    scanf("%d", &q);
    for(int i=0;i<q;++i) {
        scanf("%d", &tmp);
        x.push_back(max(global_x_ub-maxc-5, min(global_x_ub+5, tmp)));
    }
    //printf("ub: %d", global_x_ub);

    map<int,int> x2ans;
    for(int xval=global_x_ub+5;xval>=global_x_ub-maxc-5;--xval) {
        vector<int> *p_ans=nullptr, *p_ans_last=nullptr;
        p_ans_last = new vector<int>(sumc+2,0);
        (*p_ans_last)[0]=1;
        for(int i=0;i<n;++i) {
            p_ans = new vector<int>(sumc+2, 0);
            int sum_last = 0, lower_bound=xval*(i+1)+b[i];
            //printf("lb=%d, xval=%d, mul=%d, b=%d\n", lower_bound, xval, xval*(i+1), b[i]);
            for(int j=0;j<=c[i];++j) {
                sum_last = (sum_last + (*p_ans_last)[j])%M;
                (*p_ans)[j] = sum_last;
            }
            for(int j=c[i]+1;j<=sumc;++j) {
                sum_last = (sum_last - (*p_ans_last)[j-c[i]-1] + M) % M;
                sum_last = (sum_last + (*p_ans_last)[j]) % M;
                (*p_ans)[j] = sum_last;
            }
            for(int j=0;j<lower_bound && j<=sumc+1;++j)
                (*p_ans)[j]=0;
            swap(p_ans, p_ans_last);
            delete p_ans;
        }
        int cnt=0;
        for(int i=0;i<=sumc;++i)
            cnt = (cnt + (*p_ans_last)[i])%M;
        x2ans[xval] = cnt;
        delete p_ans_last;
    }

    for(int i=0;i<q;++i) {
        printf("%d\n", x2ans[x[i]]);
    }

    return 0;
}