Problem: This is the hard version of the problem. The only difference is that in this version 1 ≤ q ≤ 10^5. 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 (1 ≤ q ≤ 10^5).

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
5
-1 0 1 -100000 100000


Output


56
28
4
60
0

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<bits/stdc++.h>
using namespace std;

const int N=10005;
const int M=100005;
const int mo=1000000007;
int n,c[N],S[N],f[N],X;
int q[M],id[M],ans[M],Q;
bool cmp(int x,int y){
	return q[x]<=q[y];
}
int Calc(int X){
	memset(f,0,sizeof(f)); 
	f[0]=1;
	for (int i=1;i<=n;i++){
		for (int j=1;j<N;j++)
			f[j]=(f[j]+f[j-1])%mo;
		for (int j=N-1;j>=c[i]+1;j--)
			f[j]=(f[j]+mo-f[j-c[i]-1])%mo; 
		int s=i*X+S[i];
		for (int j=0;j<N&&j<s;j++)
			f[j]=0;
	}
	int ans=0;
	for (int j=0;j<N;j++)
		ans=(ans+f[j])%mo;
	return ans;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%d",&c[i]);
	for (int i=1;i<n;i++)
		scanf("%d",&S[i+1]);
	for (int i=1;i<=n;i++)
		S[i]+=S[i-1];
	for (int i=1;i<=n;i++)
		S[i]+=S[i-1];
	scanf("%d",&Q);
	for (int i=1;i<=Q;i++){
		scanf("%d",&q[i]);
		id[i]=i;
	}
	sort(id+1,id+Q+1,cmp);
	ans[id[1]]=Calc(q[id[1]]);
	for (int i=2;i<=Q;i++)
		if (q[id[i]]==q[id[i-1]])
			ans[id[i]]=ans[id[i-1]];
		else if (n*q[id[i]]+S[n]<0)
			ans[id[i]]=ans[id[i-1]];
		else if (n*q[id[i]]+S[n]>n*100)
			ans[id[i]]=0;
		else ans[id[i]]=Calc(q[id[i]]);
	for (int i=1;i<=Q;i++)
		printf("%d\n",ans[i]);
}