#include<bits/stdc++.h>
template <typename _Tp>void read(_Tp &x){
	char ch(getchar());bool f(false);while(!isdigit(ch))f|=ch==45,ch=getchar();
	x=ch&15,ch=getchar();while(isdigit(ch))x=x*10+(ch&15),ch=getchar();
	if(f)x=-x;
}
template <typename _Tp,typename... Args>void read(_Tp &t,Args &...args){read(t);read(args...);}
const int N=105,mod=1000000007;
template<typename _Tp1,typename _Tp2>inline void add(_Tp1 &a,_Tp2 b){(a+=b)>=mod&&(a-=mod);}
template<typename _Tp1,typename _Tp2>inline void sub(_Tp1 &a,_Tp2 b){(a-=b)<0&&(a+=mod);}
int c[N],b[N],bp[N],ap[N],dp[N][N*N],pre[N*N],ans[N+N];
int main(){
	int n;read(n);
	for(int i=1;i<=n;++i)read(c[i]),ap[i]=ap[i-1]+c[i];
	for(int i=2;i<=n;++i)read(b[i]),bp[i]=bp[i-1]+b[i];
	for(int i=2;i<=n;++i)bp[i]+=bp[i-1];
	int mn=1e5,mx=1e5;
	for(int i=1;i<=n;++i)mn=std::min(mn,-bp[i]/i),mx=std::min(mx,(ap[i]-bp[i])/i);
	--mn,++mx;
	for(int k=mn;k<=mx;++k){
		dp[0][0]=1;int s=0;
		for(int i=1;i<=n;++i){
			int lim=std::max(0,bp[i]+i*k);s+=c[i];
			pre[0]=dp[i-1][0];for(int j=1;j<=s;++j)pre[j]=pre[j-1],add(pre[j],dp[i-1][j]);
			memset(dp[i],0,lim<<2);for(int j=lim;j<=s;++j)dp[i][j]=pre[j],sub(dp[i][j],j>=c[i]+1?pre[j-c[i]-1]:0);
		}
		for(int i=0;i<=s;++i)add(ans[k-mn],dp[n][i]);
	}
	int q;read(q);
	while(q--){
		int x;read(x),x=std::min(mx,std::max(x,mn));
		printf("%d\n",ans[x-mn]);
	}
	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<bits/stdc++.h>
template <typename _Tp>void read(_Tp &x){
	char ch(getchar());bool f(false);while(!isdigit(ch))f|=ch==45,ch=getchar();
	x=ch&15,ch=getchar();while(isdigit(ch))x=x*10+(ch&15),ch=getchar();
	if(f)x=-x;
}
template <typename _Tp,typename... Args>void read(_Tp &t,Args &...args){read(t);read(args...);}
const int N=105,mod=1000000007;
template<typename _Tp1,typename _Tp2>inline void add(_Tp1 &a,_Tp2 b){(a+=b)>=mod&&(a-=mod);}
template<typename _Tp1,typename _Tp2>inline void sub(_Tp1 &a,_Tp2 b){(a-=b)<0&&(a+=mod);}
int c[N],b[N],bp[N],ap[N],dp[N][N*N],pre[N*N],ans[N+N];
int main(){
	int n;read(n);
	for(int i=1;i<=n;++i)read(c[i]),ap[i]=ap[i-1]+c[i];
	for(int i=2;i<=n;++i)read(b[i]),bp[i]=bp[i-1]+b[i];
	for(int i=2;i<=n;++i)bp[i]+=bp[i-1];
	int mn=1e5,mx=1e5;
	for(int i=1;i<=n;++i)mn=std::min(mn,-bp[i]/i),mx=std::min(mx,(ap[i]-bp[i])/i);
	--mn,++mx;
	for(int k=mn;k<=mx;++k){
		dp[0][0]=1;int s=0;
		for(int i=1;i<=n;++i){
			int lim=std::max(0,bp[i]+i*k);s+=c[i];
			pre[0]=dp[i-1][0];for(int j=1;j<=s;++j)pre[j]=pre[j-1],add(pre[j],dp[i-1][j]);
			memset(dp[i],0,lim<<2);for(int j=lim;j<=s;++j)dp[i][j]=pre[j],sub(dp[i][j],j>=c[i]+1?pre[j-c[i]-1]:0);
		}
		for(int i=0;i<=s;++i)add(ans[k-mn],dp[n][i]);
	}
	int q;read(q);
	while(q--){
		int x;read(x),x=std::min(mx,std::max(x,mn));
		printf("%d\n",ans[x-mn]);
	}
	return 0;
}
```