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;
const int N=109,M=10009,P=1e9+7;
int s[N],c[N],b[N],f[M],g[M],n;
map<int,int>mp;
int wk(int x){
	int i,j,k;
	for(memset(f,0,s[n]*4+4),i=f[0]=1;i<=n;++i){
		for(g[0]=f[0],j=1;j<=s[i];++j)if(g[j]=g[j-1]+f[j],g[j]>=P)g[j]-=P;
		for(j=0,k=x*i+b[i];j<=s[i];++j)if(j<k)f[j]=0;else if(f[j]=g[j]-(j-c[i]<1?0:g[j-c[i]-1]),f[j]<0)f[j]+=P;
	}
	for(i=j=0;i<=s[n];++i)j=(j+f[i])%P;
	return j;
}
int main(){
	int q,i,mn=P,mx=P,ms;
	for(i=1,scanf("%d",&n);i<=n;++i)scanf("%d",c+i);
	for(i=2;i<=n;++i)scanf("%d",b+i),b[i]+=b[i-1];
	for(i=2;i<=n;++i)b[i]+=b[i-1];
	for(i=1;i<=n;++i)mn=min(mn,-b[i]/i),s[i]=s[i-1]+c[i],mx=min(mx,(s[i]-b[i])/i);
	for(ms=wk(mn-=2),mx+=2,scanf("%d",&q);q--;){
		scanf("%d",&i);
		if(i<mn)printf("%d\n",ms);
		else if(i>mx)puts("0");
		else if(mp.count(i))printf("%d\n",mp[i]);
		else printf("%d\n",mp[i]=wk(i));
	}
	return 0;
}