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 CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1e9+7;
inline int addmod(int x)
{
	return x>=mod?x-mod:x;
}
inline int submod(int x)
{
	return x<0?x+mod:x; 
}
int n,m,q,b[105],c[105],sum[105],f[105][10005],qans[1005],pn;
int solve(int x)
{
	memset(f,0,sizeof(f));
	f[0][0]=1;
	int nw=0,ans=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<=nw;j++)
		{
			f[i][j]=addmod(f[i][j]+f[i-1][j]);
			f[i][j+c[i]+1]=submod(f[i][j+c[i]+1]-f[i-1][j]);
		}
		nw+=c[i];
		for(int j=1;j<=nw;j++)
			f[i][j]=addmod(f[i][j]+f[i][j-1]);
		for(int j=0;j<x*i+sum[i];j++)
			f[i][j]=0;
	}
	for(int i=0;i<=nw;i++)
		ans=addmod(ans+f[n][i]);
	return ans;
}
int main()
{
	scanf("%d",&n);
	int mul=1;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&c[i]);
		m=max(m,c[i]);
		mul=1ll*mul*(c[i]+1)%mod;
	}
	for(int i=1;i<n;i++)
		scanf("%d",&b[i]);
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)
			sum[i]+=(i-j)*b[j];
	pn=(-sum[n])/n;
	for(int i=0;i<=m;i++)
		qans[i]=solve(i+pn);
	scanf("%d",&q);
	for(int i=1;i<=q;i++)
	{
		int x;
		scanf("%d",&x);
		if(x<pn) printf("%d\n",mul);
		else if(x>pn+m) printf("0\n");
		else printf("%d\n",qans[x-pn]);
	}
	return 0;
}