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 ll long long
#define sz(x) ((int)x.size())
#define fi first
#define se second
#define pii pair<int,int>
#define pk push_back
const int maxn=1e2+5;
const int mo=1e9+7;
const int Max=1e5;
const int maxm=2e5+5;
using namespace std;
int rd(){
	int x=0,f=1;char c=getchar();
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))x=(x<<3)+(x<<1)+c-'0',c=getchar();
	return x*f;
}
int n,b[maxn],c[maxn],q,lim=1,L,R,m,ans[maxm],all;
int f[maxm],g[maxm];
int sum_b[maxn];
int pls(int x,int y){return x+y>=mo?x+y-mo:x+y;}
int dec(int x,int y){return x-y<0?x-y+mo:x-y;}
int solve(int x){
	int res=0;
	for(int i=1;i<=all;i++)f[i]=0;f[0]=1;
	for(int i=1;i<=n;i++){
		for(int j=0;j<=all;j++)g[j]=pls(f[j],(j?g[j-1]:0)),f[j]=0;
		for(int j=max(0,sum_b[i]+x*i);j<=all;j++)f[j]=dec(g[j],(j>c[i]?g[j-c[i]-1]:0));
	}
	for(int i=0;i<=all;i++)res=pls(res,f[i]);
	return res;
}
int main(){
	n=rd();
	for(int i=1;i<=n;i++)c[i]=rd(),lim=1LL*lim*(c[i]+1)%mo,m=max(m,c[i]),all+=c[i];
	for(int i=1;i<n;i++)b[i]=rd();
	q=rd();
	for(int i=1;i<=n;i++)for(int j=1;j<i;j++)sum_b[i]+=(i-j)*b[j];
	L=0,R=Max;
	for(int i=1;i<=n;i++)L=min(L,-((sum_b[i]-1)/i+1)),R=min(R,(n*m-sum_b[i]-1)/i+1);L=max(L,-Max);
	for(int i=L;i<=R;i++)ans[i+Max]=solve(i);
	for(int x;q;q--){
		x=rd();
		if(x<L)printf("%d\n",lim);
		else if(x>R)puts("0");
		else printf("%d\n",ans[x+Max]);
	}
	return 0;
}
