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;
#define mp make_pair
#define pb push_back
#define x first
#define y second
typedef pair<int,int> pii;
typedef long long ll;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x,T y){x<y?x=y:T();}
template <typename T> void chkmin(T &x,T y){y<x?x=y:T();}
template <typename T> void readint(T &x)
{
	x=0;int f=1;char c;
	for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-1;
	for(;isdigit(c);c=getchar())x=x*10+(c-'0');
	x*=f;
}
const int MOD=1000000007;
inline int dmy(int x){return x>=MOD?x-MOD:x;}
inline void inc(int &x,int y){x=dmy(x+y);}
int qmi(int x,int y)
{
	int ans=1;
	for(;y;y>>=1,x=1ll*x*x%MOD)
		if(y&1)ans=1ll*ans*x%MOD;
	return ans;
}
const int MAXN=105,MAXM=200005,M=100001;

int n,c[MAXN],b[MAXN];
int f[MAXN][MAXN*MAXN];
int res[MAXM];

int main()
{
	#ifdef LOCAL
	freopen("code.in","r",stdin);
//	freopen("code.out","w",stdout);
	#endif
	readint(n);
	int mulc=1;
	for(int i=1;i<=n;++i)readint(c[i]),mulc=1ll*mulc*(c[i]+1)%MOD;
	for(int i=2;i<=n;++i)readint(b[i]),b[i]+=b[i-1];
	for(int x=-M;x<=M;++x)
	{
		int sum=0,pre=0;
		bool fl1=0,fl2=0;
		for(int i=1;i<=n;++i)
		{
			sum+=b[i]+x,
			pre+=c[i];
			if(pre-sum<0)fl1=1;
			if(sum>0)fl2=1;
		}
		if(fl1)continue;
		if(!fl2){res[x+M]=mulc;continue;}
		memset(f,0,sizeof(f));
		sum=0,pre=0;
		f[0][0]=1;
		for(int i=1;i<=n;++i)
		{
			for(int j=max(sum,0);j<=pre;++j)
				inc(f[i][j],f[i-1][j]),
				inc(f[i][j+c[i]+1],MOD-f[i-1][j]);
			sum+=b[i]+x;
			pre+=c[i];
			for(int j=1;j<=pre;++j)inc(f[i][j],f[i][j-1]);
		}
		for(int j=max(sum,0);j<=pre;++j)
			inc(res[x+M],f[n][j]);
	}
	int Q;
	readint(Q);
	while(Q--)
	{
		int x;readint(x);
		printf("%d\n",res[x+M]);
	}
	return 0;
}