#include<iostream>
#include<map>
#include<cstring>
#define int long long
using namespace std;
const int N=105,p=1e9+7;
int f[N*N],sum[N*N],b[N],sb[N],c[N],l,r,all=1,n;
int ask(int x)
{
	int now=x+sb[1];
	memset(f,0,sizeof f);
	memset(sum,0,sizeof sum);
	for(int i=max(0ll,now);i<=c[1];i++)
	f[i]=1;
	sum[0]=f[0];
	for(int j=1;j<=10000;j++)
	sum[j]=(sum[j-1]+f[j])%p;
	for(int i=2;i<=n;i++)
	{
		memset(f,0,sizeof f);
		now=x*i+sb[i];
		for(int j=max(0ll,now);j<=10000;j++)
		if(j-c[i]-1>=0)
		f[j]=((sum[j]-sum[j-c[i]-1])%p+p)%p;
		else f[j]=sum[j];
		sum[0]=f[0];
		for(int j=1;j<=10000;j++)
		sum[j]=(sum[j-1]+f[j])%p;
	}
	return sum[10000];
}
map<int,int> ans;
signed main()
{
	cin>>n;
	int m=0;
	for(int i=1;i<=n;i++)
	cin>>c[i],m=max(m,c[i]),all=all*(c[i]+1)%p;
	for(int i=1;i<n;i++)
	cin>>b[i];
	for(int i=1;i<=n;i++)
	for(int j=1;j<i;j++)
	sb[i]+=(i-j)*b[j];
	l=r=2e9;
	for(int i=1;i<=n;i++)
	r=min(r,(i*m-sb[i])/i),l=min(l,-sb[i]/i);
	int q;
	cin>>q;
	for(int i=l;i<=r;i++)
	ans[i]=ask(i);
	while(q--)
	{
		int x;
		cin>>x;
		if(x<l) cout<<all<<endl;
		else if(x>r) cout<<0<<endl;
		else cout<<ans[x]<<endl;
	}
	return 0;
}
### Prompt

Create a solution in CPP for 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<iostream>
#include<map>
#include<cstring>
#define int long long
using namespace std;
const int N=105,p=1e9+7;
int f[N*N],sum[N*N],b[N],sb[N],c[N],l,r,all=1,n;
int ask(int x)
{
	int now=x+sb[1];
	memset(f,0,sizeof f);
	memset(sum,0,sizeof sum);
	for(int i=max(0ll,now);i<=c[1];i++)
	f[i]=1;
	sum[0]=f[0];
	for(int j=1;j<=10000;j++)
	sum[j]=(sum[j-1]+f[j])%p;
	for(int i=2;i<=n;i++)
	{
		memset(f,0,sizeof f);
		now=x*i+sb[i];
		for(int j=max(0ll,now);j<=10000;j++)
		if(j-c[i]-1>=0)
		f[j]=((sum[j]-sum[j-c[i]-1])%p+p)%p;
		else f[j]=sum[j];
		sum[0]=f[0];
		for(int j=1;j<=10000;j++)
		sum[j]=(sum[j-1]+f[j])%p;
	}
	return sum[10000];
}
map<int,int> ans;
signed main()
{
	cin>>n;
	int m=0;
	for(int i=1;i<=n;i++)
	cin>>c[i],m=max(m,c[i]),all=all*(c[i]+1)%p;
	for(int i=1;i<n;i++)
	cin>>b[i];
	for(int i=1;i<=n;i++)
	for(int j=1;j<i;j++)
	sb[i]+=(i-j)*b[j];
	l=r=2e9;
	for(int i=1;i<=n;i++)
	r=min(r,(i*m-sb[i])/i),l=min(l,-sb[i]/i);
	int q;
	cin>>q;
	for(int i=l;i<=r;i++)
	ans[i]=ask(i);
	while(q--)
	{
		int x;
		cin>>x;
		if(x<l) cout<<all<<endl;
		else if(x>r) cout<<0<<endl;
		else cout<<ans[x]<<endl;
	}
	return 0;
}
```