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.

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

const int maxn=105;
const int mod=1e9+7;
const int R=1e5;
int n,t,q;
int b[maxn],c[maxn],s[maxn];

void add(int &a,int b){
	a+=b;
	if(a>=mod)a-=mod;
}
void sub(int &a,int b){
	a-=b;
	if(a<0)a+=mod;
}

class Func{
public:
	vector<pair<int,int> >val;
	Func(){
		val.clear();
	}
	void init(){
		val.push_back(make_pair(R,1)); 
	}
	void pop(int p){
		while(int(val.size())>=2&&val[int(val.size())-2].first>=p)
			val.pop_back();
		if(!val.empty())val.back().first=min(val.back().first,p);
	}
	Func operator+(const Func &o)const{
		Func res;
		int i=0,oi=0,sz=int(val.size()),osz=int(o.val.size());
		while(i<sz||oi<osz){
			int npos=R,nval=0;
			if(i<sz)npos=min(npos,val[i].first),add(nval,val[i].second);
			if(oi<osz)npos=min(npos,o.val[oi].first),add(nval,o.val[oi].second);
			res.val.push_back(make_pair(npos,nval));
			if(i<sz&&val[i].first==npos)i++;
			if(oi<osz&&o.val[oi].first==npos)oi++;
		}
		return res;
	}
	Func operator-(const Func &o)const{
		Func res;
		int i=0,oi=0,sz=int(val.size()),osz=int(o.val.size());
		while(i<sz||oi<osz){
			int npos=R,nval=0;
			if(i<sz)npos=min(npos,val[i].first),add(nval,val[i].second);
			if(oi<osz)npos=min(npos,o.val[oi].first),sub(nval,o.val[oi].second);
			res.val.push_back(make_pair(npos,nval));
			if(i<sz&&val[i].first==npos)i++;
			if(oi<osz&&o.val[oi].first==npos)oi++;
		}
		return res;
	}
	int query(int x){
		int p=lower_bound(val.begin(),val.end(),make_pair(x,-1))-val.begin();
		return p<int(val.size())?val[p].second:0;
	}
}dp[maxn*maxn],ndp[maxn*maxn],ans;

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&c[i]);
	for(int i=2;i<=n;i++)scanf("%d",&b[i]);
	dp[0].init();
	for(int i=1;i<=n;i++){
		s[i]=s[i-1]+c[i];
		b[i]+=b[i-1];
		t+=b[i];
		Func cur; 
		for(int j=0;j<=s[i];j++){
			if(j<=s[i-1])cur=cur+dp[j];
			if(j-c[i]-1>=0)cur=cur-dp[j-c[i]-1];
			ndp[j]=cur;
			int x=(j-t)/i;
			while(j>=t+i*(x+1))x++;
			while(j<t+i*x)x--;
			ndp[j].pop(x);
		}
		for(int j=0;j<=s[i];j++)
			dp[j]=ndp[j];
	}
	for(int j=0;j<=s[n];j++)
		ans=ans+dp[j];
	scanf("%d",&q);
	while(q--){
		int x;
		scanf("%d",&x);
		printf("%d\n",ans.query(x));
	}
	return 0;
}