#include<bits/stdc++.h>
using namespace std;
const int N=105,mod=1e9+7;
int n,q;
int b[N],c[N],sb[N],sc[N];
int f[N][N*N];
map<int,int> mp;
inline void dec(int &x,int y){
	x-=y;if(x<0) x+=mod;
}
inline void add(int &x,int y){
	x+=y;if(x>=mod) x-=mod;
}
inline int calc(int x){
	memset(f,0,sizeof(f));
	for(int i=0;i<=sc[1];i++) f[0][i]=1;
	for(int i=1;i<=n;i++){
		for(int j=max(i*x+sb[i],0);j<=sc[i];j++){
			f[i][j]=f[i-1][j];
			if(j-c[i]-1>=0) dec(f[i][j],f[i-1][j-c[i]-1]); 
		}
		for(int j=1;j<=sc[i+1];j++) add(f[i][j],f[i][j-1]);
	}
	return f[n][sc[n]];
}
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin>>n;
	int tot=1;
	for(int i=1;i<=n;i++) cin>>c[i],sc[i]=sc[i-1]+c[i],tot=1ll*tot*(c[i]+1)%mod;sc[n+1]=sc[n];
	for(int i=1;i<n;i++) cin>>b[i];
	for(int i=2;i<=n;i++) sb[i]=sb[i-1]+b[i-1];
	for(int i=1;i<=n;i++) sb[i]=sb[i]+sb[i-1];
	cin>>q;
	while(q--){
		int x;cin>>x;
		if(x>100) cout<<"0\n";
		else if(x<-5000) cout<<tot<<'\n';
		else if(mp.find(x)!=mp.end()) cout<<mp[x]<<'\n';
		else cout<<(mp[x]=calc(x))<<'\n';
	}
	return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;
const int N=105,mod=1e9+7;
int n,q;
int b[N],c[N],sb[N],sc[N];
int f[N][N*N];
map<int,int> mp;
inline void dec(int &x,int y){
	x-=y;if(x<0) x+=mod;
}
inline void add(int &x,int y){
	x+=y;if(x>=mod) x-=mod;
}
inline int calc(int x){
	memset(f,0,sizeof(f));
	for(int i=0;i<=sc[1];i++) f[0][i]=1;
	for(int i=1;i<=n;i++){
		for(int j=max(i*x+sb[i],0);j<=sc[i];j++){
			f[i][j]=f[i-1][j];
			if(j-c[i]-1>=0) dec(f[i][j],f[i-1][j-c[i]-1]); 
		}
		for(int j=1;j<=sc[i+1];j++) add(f[i][j],f[i][j-1]);
	}
	return f[n][sc[n]];
}
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin>>n;
	int tot=1;
	for(int i=1;i<=n;i++) cin>>c[i],sc[i]=sc[i-1]+c[i],tot=1ll*tot*(c[i]+1)%mod;sc[n+1]=sc[n];
	for(int i=1;i<n;i++) cin>>b[i];
	for(int i=2;i<=n;i++) sb[i]=sb[i-1]+b[i-1];
	for(int i=1;i<=n;i++) sb[i]=sb[i]+sb[i-1];
	cin>>q;
	while(q--){
		int x;cin>>x;
		if(x>100) cout<<"0\n";
		else if(x<-5000) cout<<tot<<'\n';
		else if(mp.find(x)!=mp.end()) cout<<mp[x]<<'\n';
		else cout<<(mp[x]=calc(x))<<'\n';
	}
	return 0;
}
```