#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<cassert>
#include<set>
#include<map>
using namespace std;
#define pb push_back
#define mp make_pair
#define rep(i,n) for(int i=1;i<=n;i++)
typedef long long LL;
const int MOD=(int)1e9+7;
const int N=110;
int f[N][N*N],ssb[N],sc[N],b[N],c[N],n,q,ans[N];
void work(int&ans,int x)
{
	memset(f,0,sizeof(f));
	rep(i,n)
	{
		if(i==1)for(int j=max(0,x);j<=c[1];j++)f[1][j]=1;
		else
		{
			rep(j,sc[i])(f[i-1][j]+=f[i-1][j-1])%=MOD;
			for(int j=max(0,i*x+ssb[i]);j<=sc[i];j++)f[i][j]=(f[i-1][j]-(j-c[i]-1>=0?f[i-1][j-c[i]-1]:0)+MOD)%MOD;
		}
	}
	ans=0;
	for(int i=0;i<=sc[n];i++)(ans+=f[n][i])%=MOD;
}
int main()
{
	scanf("%d",&n);
	rep(i,n)scanf("%d",&c[i]),sc[i]=sc[i-1]+c[i];
	rep(i,n-1)scanf("%d",&b[i+1]),ssb[i+1]=ssb[i]+b[i+1];
	rep(i,n)ssb[i]+=ssb[i-1];
	int L=(-ssb[n])/n,R=(sc[n]-ssb[n])/n;
	rep(i,n-1)L=min(L,(-ssb[i])/i),R=min(R,(sc[i]-ssb[i])/i);
	L=max(L,-(int)1e5);R=min(R,(int)1e5);
	for(int i=L;i<=R;i++)work(ans[i-L+1],i);
	scanf("%d",&q);
	int pw2=1;
	rep(i,n)pw2=1ll*pw2*(c[i]+1)%MOD;
	rep(Q,q)
	{
		int x;scanf("%d",&x);
		if(x>R)puts("0");
		else if(x<L)printf("%d\n",pw2);
		else printf("%d\n",ans[x-L+1]);
	}
    return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<cassert>
#include<set>
#include<map>
using namespace std;
#define pb push_back
#define mp make_pair
#define rep(i,n) for(int i=1;i<=n;i++)
typedef long long LL;
const int MOD=(int)1e9+7;
const int N=110;
int f[N][N*N],ssb[N],sc[N],b[N],c[N],n,q,ans[N];
void work(int&ans,int x)
{
	memset(f,0,sizeof(f));
	rep(i,n)
	{
		if(i==1)for(int j=max(0,x);j<=c[1];j++)f[1][j]=1;
		else
		{
			rep(j,sc[i])(f[i-1][j]+=f[i-1][j-1])%=MOD;
			for(int j=max(0,i*x+ssb[i]);j<=sc[i];j++)f[i][j]=(f[i-1][j]-(j-c[i]-1>=0?f[i-1][j-c[i]-1]:0)+MOD)%MOD;
		}
	}
	ans=0;
	for(int i=0;i<=sc[n];i++)(ans+=f[n][i])%=MOD;
}
int main()
{
	scanf("%d",&n);
	rep(i,n)scanf("%d",&c[i]),sc[i]=sc[i-1]+c[i];
	rep(i,n-1)scanf("%d",&b[i+1]),ssb[i+1]=ssb[i]+b[i+1];
	rep(i,n)ssb[i]+=ssb[i-1];
	int L=(-ssb[n])/n,R=(sc[n]-ssb[n])/n;
	rep(i,n-1)L=min(L,(-ssb[i])/i),R=min(R,(sc[i]-ssb[i])/i);
	L=max(L,-(int)1e5);R=min(R,(int)1e5);
	for(int i=L;i<=R;i++)work(ans[i-L+1],i);
	scanf("%d",&q);
	int pw2=1;
	rep(i,n)pw2=1ll*pw2*(c[i]+1)%MOD;
	rep(Q,q)
	{
		int x;scanf("%d",&x);
		if(x>R)puts("0");
		else if(x<L)printf("%d\n",pw2);
		else printf("%d\n",ans[x-L+1]);
	}
    return 0;
}
```