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 ls i<<1
#define rs i<<1|1
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long ll;
typedef vector<int> vct;
typedef pair<int,int> pii;
const int N=101,M=1E5+1;
const int inf=1<<30;
ll read()
{
    ll a=0;
    char b=1,c;
    do if((c=getchar())==45)b=-1;while(c<48||c>57);
    do a=(a<<3)+(a<<1)+(c&15);while((c=getchar())>47&&c<58);
    return a*b;
}
void write(ll x,char c)
{
    if(x<0)putchar(45),x=-x;
    char a[20],s=0;
    do a[++s]=x%10|48;while(x/=10);
    do putchar(a[s]);while(--s);
    putchar(c);
}
int a[N],b[N],sum[N];
int f[N][N*N],ANS[2][M];
int main()
{
    const int p=1E9+7;
    int n=read();
    int all=1;
    for(int i=1;i<=n;++i)
    {
        a[i]=read();
        all=all*(a[i]+1LL)%p;
        sum[i]=sum[i-1]+a[i];
    }
    sum[n+1]=sum[n];
    for(int i=2;i<=n;++i)b[i]=b[i-1]+read();
    for(int i=2;i<=n;++i)b[i]+=b[i-1];
    int *ans=ANS[1];
    for(int i=-M;i<M;++i)
    {
        if(n*i+b[n]>n*100)continue;
        if(n*i+b[n]<=0)ans[i]=all;
        else
        {
            memset(f,0,sizeof f);
            for(int k=0;k<=sum[1];++k)f[0][k]=1;
            for(int j=1;j<=n;++j)
            {
                int y=max(j*i+b[j],0);
                for(int k=y;k<=sum[j];++k)
                {
                    f[j][k]=f[j-1][k];
                    if(k>a[j])f[j][k]=(f[j][k]-f[j-1][k-a[j]-1]+p)%p;
                }
                for(int k=y+1;k<=sum[j+1];++k)f[j][k]=(f[j][k]+f[j][k-1])%p;
            }
            ans[i]=f[n][sum[n]];
        }
    }
    for(int m=read();m--;)write(ans[read()],'\n');
}
