#include<cstdio>
#include<algorithm>
using namespace std;

const int md=1000000007;
int f[101][10001];
int a[101],b[101],x[100001],ans[100001];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=2;i<=n;i++){
		scanf("%d",&b[i]);
		b[i]+=b[i-1];
	}
	for(int i=2;i<=n;i++)
		b[i]+=b[i-1];
	int q;
	scanf("%d",&q);
	for(int i=1;i<=q;i++)
		scanf("%d",&x[i]);
	int mn=2e9,mx=2e9;
	for(int i=1,z=0;i<=n;i++){
		mn=min(mn,-b[i]/i);
		z+=a[i];
		mx=min(mx,(z-b[i])/i+1);
	}
	mn--;
	f[0][0]=1;
	for(int j=1;j<=10000;j++)
		f[0][j]=(f[0][j]+f[0][j-1])%md;
	for(int o=mn;o<=mx;o++){
		for(int i=1;i<=n;i++){
			for(int j=0;j<=10000;j++){
				if(j-b[i]>=i*o){
					int k=min(a[i],j);
					f[i][j]=(f[i-1][j]-(j-k-1==-1?0:f[i-1][j-k-1])+md)%md;
				}
				else f[i][j]=0;
			}
			for(int j=1;j<=10000;j++)
				f[i][j]=(f[i][j]+f[i][j-1])%md;
		}
		ans[o-mn]=f[n][10000];
	}
	for(int i=1;i<=q;i++){
		x[i]=max(x[i],mn);
		x[i]=min(x[i],mx);
		printf("%d\n",ans[x[i]-mn]);
	}
}

### Prompt

Please create a solution in Cpp to the following 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.

### Response

```cpp
#include<cstdio>
#include<algorithm>
using namespace std;

const int md=1000000007;
int f[101][10001];
int a[101],b[101],x[100001],ans[100001];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=2;i<=n;i++){
		scanf("%d",&b[i]);
		b[i]+=b[i-1];
	}
	for(int i=2;i<=n;i++)
		b[i]+=b[i-1];
	int q;
	scanf("%d",&q);
	for(int i=1;i<=q;i++)
		scanf("%d",&x[i]);
	int mn=2e9,mx=2e9;
	for(int i=1,z=0;i<=n;i++){
		mn=min(mn,-b[i]/i);
		z+=a[i];
		mx=min(mx,(z-b[i])/i+1);
	}
	mn--;
	f[0][0]=1;
	for(int j=1;j<=10000;j++)
		f[0][j]=(f[0][j]+f[0][j-1])%md;
	for(int o=mn;o<=mx;o++){
		for(int i=1;i<=n;i++){
			for(int j=0;j<=10000;j++){
				if(j-b[i]>=i*o){
					int k=min(a[i],j);
					f[i][j]=(f[i-1][j]-(j-k-1==-1?0:f[i-1][j-k-1])+md)%md;
				}
				else f[i][j]=0;
			}
			for(int j=1;j<=10000;j++)
				f[i][j]=(f[i][j]+f[i][j-1])%md;
		}
		ans[o-mn]=f[n][10000];
	}
	for(int i=1;i<=q;i++){
		x[i]=max(x[i],mn);
		x[i]=min(x[i],mx);
		printf("%d\n",ans[x[i]-mn]);
	}
}

```