#include <iostream>
#include <limits.h>
#include <cstring>
using namespace std;
const int N=105,mo=1000000007;
int n;
int c[N],b[N];
int sc[N],sb0[N],sb1[N],h[N];
int lim;
int f[N][N*N],g[N*N];
int ans[N];
int solve(){
	memset(f,0,sizeof f);
	f[0][0]=1;
	for (int i=1;i<=n;++i){
		long long ps=0;
		for (int s=0;s<=sc[i];++s)
			g[s]=(ps+=f[i-1][s])%=mo;
		for (int s=sc[i];s>c[i];--s)
			g[s]=(g[s]-g[s-c[i]-1]+mo)%mo;
		int t=max(lim*i+h[i],0);
		for (int s=t;s<=sc[i];++s)
			f[i][s]=g[s];
	}
	long long ans=0;
	for (int s=0;s<=sc[n];++s)
		ans+=f[n][s];
	ans%=mo;
	return ans;
}
int main(){
	cin>>n;
	for (int i=1;i<=n;++i){
		cin>>c[i];
		sc[i]=sc[i-1]+c[i];
	}
	for (int i=1;i<n;++i)
		cin>>b[i];
	int mx=INT_MIN;
	for (int i=1;i<=n;++i){
		sb0[i]=sb0[i-1]+b[i];
		sb1[i]=sb1[i-1]+b[i]*i;
		h[i]=sb0[i]*i-sb1[i];
		mx=max(mx,(h[i]+i-1)/i);
	}
	for (int i=0;i<=101;++i)
		lim=i-mx,ans[i]=solve();
	int Q;
	cin>>Q;
	while (Q--){
		cin>>lim;
		lim=min(max(lim,0-mx),101-mx);
		cout<<ans[lim+mx]<<endl;
	}
	return 0;
}

### Prompt

Please formulate a CPP solution 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 <iostream>
#include <limits.h>
#include <cstring>
using namespace std;
const int N=105,mo=1000000007;
int n;
int c[N],b[N];
int sc[N],sb0[N],sb1[N],h[N];
int lim;
int f[N][N*N],g[N*N];
int ans[N];
int solve(){
	memset(f,0,sizeof f);
	f[0][0]=1;
	for (int i=1;i<=n;++i){
		long long ps=0;
		for (int s=0;s<=sc[i];++s)
			g[s]=(ps+=f[i-1][s])%=mo;
		for (int s=sc[i];s>c[i];--s)
			g[s]=(g[s]-g[s-c[i]-1]+mo)%mo;
		int t=max(lim*i+h[i],0);
		for (int s=t;s<=sc[i];++s)
			f[i][s]=g[s];
	}
	long long ans=0;
	for (int s=0;s<=sc[n];++s)
		ans+=f[n][s];
	ans%=mo;
	return ans;
}
int main(){
	cin>>n;
	for (int i=1;i<=n;++i){
		cin>>c[i];
		sc[i]=sc[i-1]+c[i];
	}
	for (int i=1;i<n;++i)
		cin>>b[i];
	int mx=INT_MIN;
	for (int i=1;i<=n;++i){
		sb0[i]=sb0[i-1]+b[i];
		sb1[i]=sb1[i-1]+b[i]*i;
		h[i]=sb0[i]*i-sb1[i];
		mx=max(mx,(h[i]+i-1)/i);
	}
	for (int i=0;i<=101;++i)
		lim=i-mx,ans[i]=solve();
	int Q;
	cin>>Q;
	while (Q--){
		cin>>lim;
		lim=min(max(lim,0-mx),101-mx);
		cout<<ans[lim+mx]<<endl;
	}
	return 0;
}

```