#include<bits/stdc++.h>
using namespace std;
namespace IO{
	char pool[1<<20],*p1=pool,*p2=pool;
	#define gc (p1==p2&&(p2=(p1=pool)+fread(pool,1,1<<20,stdin),p1==p2)?EOF:*p1++)
	int R(){
		int x=0,y=1;char ch=gc;
		while(ch<'0'||ch>'9'){if(ch=='-')y=-1;ch=gc;}
		while(ch>='0'&&ch<='9')x=x*10+(ch^48),ch=gc;
		return x*y;
	}
}
const int N=105,M=1e4,P=1e9+7;
int n,Q,ways=1,m,lb,rb=2e9,c[N],s[N];
int f[M+5],g[M+5];map<int,int>ans;
int xxx(int x){
    for(int i=0;i<=M;i++)g[i]=1;
    for(int i=1;i<=n;i++){
        memset(f,0,sizeof f);
        for(int j=max(0,i*x+s[i]);j<=M;j++)
        f[j]=g[j]-(j>c[i]?g[j-c[i]-1]:0),f[j]+=f[j]<0?P:0;
        memset(g,0,sizeof g);
        for(int j=0;j<=M;j++)g[j]=(j?g[j-1]:0)+f[j],g[j]-=g[j]<P?0:P;
    }
	return g[M];
}
int main(){
    n=IO::R();
    for(int i=1;i<=n;i++){
        c[i]=IO::R();m=max(c[i],m);
        ways=1ll*ways*(c[i]+1)%P;
    }
    for(int i=1;i<n;i++){
        int x=IO::R();
        for(int j=i+1;j<=n;j++)
        s[j]+=1ll*(j-i)*x%P,s[j]-=s[j]<P?0:P;
    }
	for(int i=1;i<=n;i++)
    lb=min(lb,-((s[i]-1)/i+1)),rb=min(rb,(n*m-s[i]-1)/i+1);
	for(int i=lb;i<=rb;i++)ans[i]=xxx(i);
	for(Q=IO::R();Q--;){
	    int x=IO::R();
	    if(x<lb)printf("%d\n",ways);
	    else if(x>rb)puts("0");
	    else printf("%d\n",ans[x]);
    }
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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<bits/stdc++.h>
using namespace std;
namespace IO{
	char pool[1<<20],*p1=pool,*p2=pool;
	#define gc (p1==p2&&(p2=(p1=pool)+fread(pool,1,1<<20,stdin),p1==p2)?EOF:*p1++)
	int R(){
		int x=0,y=1;char ch=gc;
		while(ch<'0'||ch>'9'){if(ch=='-')y=-1;ch=gc;}
		while(ch>='0'&&ch<='9')x=x*10+(ch^48),ch=gc;
		return x*y;
	}
}
const int N=105,M=1e4,P=1e9+7;
int n,Q,ways=1,m,lb,rb=2e9,c[N],s[N];
int f[M+5],g[M+5];map<int,int>ans;
int xxx(int x){
    for(int i=0;i<=M;i++)g[i]=1;
    for(int i=1;i<=n;i++){
        memset(f,0,sizeof f);
        for(int j=max(0,i*x+s[i]);j<=M;j++)
        f[j]=g[j]-(j>c[i]?g[j-c[i]-1]:0),f[j]+=f[j]<0?P:0;
        memset(g,0,sizeof g);
        for(int j=0;j<=M;j++)g[j]=(j?g[j-1]:0)+f[j],g[j]-=g[j]<P?0:P;
    }
	return g[M];
}
int main(){
    n=IO::R();
    for(int i=1;i<=n;i++){
        c[i]=IO::R();m=max(c[i],m);
        ways=1ll*ways*(c[i]+1)%P;
    }
    for(int i=1;i<n;i++){
        int x=IO::R();
        for(int j=i+1;j<=n;j++)
        s[j]+=1ll*(j-i)*x%P,s[j]-=s[j]<P?0:P;
    }
	for(int i=1;i<=n;i++)
    lb=min(lb,-((s[i]-1)/i+1)),rb=min(rb,(n*m-s[i]-1)/i+1);
	for(int i=lb;i<=rb;i++)ans[i]=xxx(i);
	for(Q=IO::R();Q--;){
	    int x=IO::R();
	    if(x<lb)printf("%d\n",ways);
	    else if(x>rb)puts("0");
	    else printf("%d\n",ans[x]);
    }
	return 0;
}
```