#include <bits/stdc++.h>
using namespace std;
#define N 105
#define M 200005
#define MOD 1000000007
const int lim=1e5;
int n,m,L=lim,R=lim,a[N],b[N],ans[M],dp[N*N],s[N*N];
int f(int x,int y) {return x>0?x/y:-(-x-1)/y-1;}
int slv(int x)
{
	int res=0;for(int i=0;i<=a[n];++i) dp[i]=0;dp[0]=1;
	for(int i=1,t;i<=n;++i)
	{
		for(int j=0;j<=a[n];++j)
			s[j]=(dp[j]+(j?s[j-1]:0))%MOD,dp[j]=0;
		for(int j=max(b[i]+x*i,0);j<=a[n];++j)
			t=a[i]-a[i-1],dp[j]=(s[j]-(j>t?s[j-t-1]:0))%MOD;
	}for(int i=0;i<=a[n];++i) res=(res+dp[i])%MOD;
	return (res+MOD)%MOD;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i) scanf("%d",&a[i]),a[i]+=a[i-1];
	for(int i=2;i<=n;++i) scanf("%d",&b[i]),b[i]+=b[i-1];
	scanf("%d",&m);for(int i=2;i<=n;++i) b[i]+=b[i-1];
	for(int i=1;i<=n;++i)
		L=min(L,f(-b[i],i)),R=min(R,f(a[i]-b[i],i));
	L=max(L,-lim);for(int i=L;i<=R;++i) ans[lim+i]=slv(i);
	for(int i=1,x;i<=m;++i)
		scanf("%d",&x),printf("%d\n",ans[lim+max(x,L)]);return 0;
}
### Prompt

In Cpp, your task is to solve 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;
#define N 105
#define M 200005
#define MOD 1000000007
const int lim=1e5;
int n,m,L=lim,R=lim,a[N],b[N],ans[M],dp[N*N],s[N*N];
int f(int x,int y) {return x>0?x/y:-(-x-1)/y-1;}
int slv(int x)
{
	int res=0;for(int i=0;i<=a[n];++i) dp[i]=0;dp[0]=1;
	for(int i=1,t;i<=n;++i)
	{
		for(int j=0;j<=a[n];++j)
			s[j]=(dp[j]+(j?s[j-1]:0))%MOD,dp[j]=0;
		for(int j=max(b[i]+x*i,0);j<=a[n];++j)
			t=a[i]-a[i-1],dp[j]=(s[j]-(j>t?s[j-t-1]:0))%MOD;
	}for(int i=0;i<=a[n];++i) res=(res+dp[i])%MOD;
	return (res+MOD)%MOD;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i) scanf("%d",&a[i]),a[i]+=a[i-1];
	for(int i=2;i<=n;++i) scanf("%d",&b[i]),b[i]+=b[i-1];
	scanf("%d",&m);for(int i=2;i<=n;++i) b[i]+=b[i-1];
	for(int i=1;i<=n;++i)
		L=min(L,f(-b[i],i)),R=min(R,f(a[i]-b[i],i));
	L=max(L,-lim);for(int i=L;i<=R;++i) ans[lim+i]=slv(i);
	for(int i=1,x;i<=m;++i)
		scanf("%d",&x),printf("%d\n",ans[lim+max(x,L)]);return 0;
}
```