#include<bits/stdc++.h>
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define mod 1000000007
using namespace std;
int n,m;
int c[109],b[109];
int smb[109],ssb[109];
int f[109][10009],sf[109][10009],g[10009];
int pt[109],smc[109];
int mo(int x)
{
	f[0][0]=1;
	for(int i=1;i<=n;i++)
	{
		int mn=i*x+ssb[i-1],lp=0;
		ll sm=0;
		for(int j=0;j<=smc[i];j++)
		{
			sm+=f[i-1][j];
			(lp<j-c[i])&&(sm+=mod-f[i-1][lp]);
			lp+=(lp<j-c[i]);
			// printf("i:%d j:%d lp:%d sm:%d\n",i,j,lp,sm);
			f[i][j]=j>=mn?sm%mod:0;
		}
	}
	int ans=0;
	for(int i=0;i<=smc[n];i++)
		ans=(ans+f[n][i])%mod;
	return ans;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",c+i);
		smc[i]=smc[i-1]+c[i];
	}
	for(int i=1;i<=n-1;i++)
	{
		scanf("%d",b+i);
		smb[i]=smb[i-1]+b[i];
		ssb[i]=ssb[i-1]+smb[i];
	}
	map<int,int> ans;
	int lb=(-ssb[n-1])/n-5;
	int rb=(smc[n]-ssb[n-1])/n+5;
	for(int i=lb;i<=rb;i++)
		ans[i]=mo(i);
	scanf("%d",&m);
	for(int i=1;i<=m;i++)
	{
		int a;
		scanf("%d",&a);
		if(a<lb)
			printf("%d\n",ans[lb]);
		else if(a>rb)
			printf("0\n");
		else
			printf("%d\n",ans[a]);
	}
	return 0;
}
### Prompt

Your task is to create 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>
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define mod 1000000007
using namespace std;
int n,m;
int c[109],b[109];
int smb[109],ssb[109];
int f[109][10009],sf[109][10009],g[10009];
int pt[109],smc[109];
int mo(int x)
{
	f[0][0]=1;
	for(int i=1;i<=n;i++)
	{
		int mn=i*x+ssb[i-1],lp=0;
		ll sm=0;
		for(int j=0;j<=smc[i];j++)
		{
			sm+=f[i-1][j];
			(lp<j-c[i])&&(sm+=mod-f[i-1][lp]);
			lp+=(lp<j-c[i]);
			// printf("i:%d j:%d lp:%d sm:%d\n",i,j,lp,sm);
			f[i][j]=j>=mn?sm%mod:0;
		}
	}
	int ans=0;
	for(int i=0;i<=smc[n];i++)
		ans=(ans+f[n][i])%mod;
	return ans;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",c+i);
		smc[i]=smc[i-1]+c[i];
	}
	for(int i=1;i<=n-1;i++)
	{
		scanf("%d",b+i);
		smb[i]=smb[i-1]+b[i];
		ssb[i]=ssb[i-1]+smb[i];
	}
	map<int,int> ans;
	int lb=(-ssb[n-1])/n-5;
	int rb=(smc[n]-ssb[n-1])/n+5;
	for(int i=lb;i<=rb;i++)
		ans[i]=mo(i);
	scanf("%d",&m);
	for(int i=1;i<=m;i++)
	{
		int a;
		scanf("%d",&a);
		if(a<lb)
			printf("%d\n",ans[lb]);
		else if(a>rb)
			printf("0\n");
		else
			printf("%d\n",ans[a]);
	}
	return 0;
}
```