#include<bits/stdc++.h>
#define int long long
#define pb push_back
#define mp make_pair
#define x first
#define y second
#define WT int TT=read();while(TT--) 
#define NO puts("NO");
#define YES puts("YES");
using namespace std;

inline int read()
{
    char c=getchar();int x=0;bool f=0;
    for(;!isdigit(c);c=getchar())f^=!(c^45);
    for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
    if(f)x=-x;return x;
}

const int Mod=1e9+7;
const int M=110;
int dp[M][M*M],n,d[M],c[M],s[M],sc[M],sum[M][M*M],xx=1;
map<int,int>f;

void solve(int x)
{
	if (f[x]){printf("%lld\n",f[x]);return;}
	s[1]=x;
	for (int i=2;i<=n;i++)s[i]=s[i-1]+d[i-1];
	for (int i=1;i<=n;i++)s[i]+=s[i-1];
	if (s[n]<=0){printf("%lld\n",xx);return;}
	int ff=0,ss=0;
	for (int i=1;i<=n;i++)
	{
		ss+=c[i];
		if (ss<s[i])ff=1;
	}
	if (ff){puts("0");return;}
	memset(dp,0,sizeof(dp)),memset(sum,0,sizeof(sum));
	dp[0][0]=1,sum[0][0]=1;
	for (int i=1;i<=n;i++)
		for (int j=max(s[i],0ll);j<=sc[i];j++)
		{
			dp[i][j]=sum[i-1][min(j,sc[i-1])];
			if (j>c[i])
				dp[i][j]=(dp[i][j]-sum[i-1][min(sc[i-1],j-c[i]-1)]+Mod)%Mod;
			sum[i][j]=(sum[i][j-1]+dp[i][j])%Mod;
		}
	int ans=0;
	for (int i=max(s[n],0ll);i<=sc[n];i++)ans=(ans+dp[n][i])%Mod;
	cout<<(f[x]=ans)<<endl;
}

signed main()
{
	n=read();
	for (int i=1;i<=n;i++)sc[i]=c[i]=read(),sc[i]+=sc[i-1],xx=xx*(c[i]+1)%Mod;
	for (int i=1;i<n;i++)d[i]=read();
	WT{solve(read());}
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 int long long
#define pb push_back
#define mp make_pair
#define x first
#define y second
#define WT int TT=read();while(TT--) 
#define NO puts("NO");
#define YES puts("YES");
using namespace std;

inline int read()
{
    char c=getchar();int x=0;bool f=0;
    for(;!isdigit(c);c=getchar())f^=!(c^45);
    for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
    if(f)x=-x;return x;
}

const int Mod=1e9+7;
const int M=110;
int dp[M][M*M],n,d[M],c[M],s[M],sc[M],sum[M][M*M],xx=1;
map<int,int>f;

void solve(int x)
{
	if (f[x]){printf("%lld\n",f[x]);return;}
	s[1]=x;
	for (int i=2;i<=n;i++)s[i]=s[i-1]+d[i-1];
	for (int i=1;i<=n;i++)s[i]+=s[i-1];
	if (s[n]<=0){printf("%lld\n",xx);return;}
	int ff=0,ss=0;
	for (int i=1;i<=n;i++)
	{
		ss+=c[i];
		if (ss<s[i])ff=1;
	}
	if (ff){puts("0");return;}
	memset(dp,0,sizeof(dp)),memset(sum,0,sizeof(sum));
	dp[0][0]=1,sum[0][0]=1;
	for (int i=1;i<=n;i++)
		for (int j=max(s[i],0ll);j<=sc[i];j++)
		{
			dp[i][j]=sum[i-1][min(j,sc[i-1])];
			if (j>c[i])
				dp[i][j]=(dp[i][j]-sum[i-1][min(sc[i-1],j-c[i]-1)]+Mod)%Mod;
			sum[i][j]=(sum[i][j-1]+dp[i][j])%Mod;
		}
	int ans=0;
	for (int i=max(s[n],0ll);i<=sc[n];i++)ans=(ans+dp[n][i])%Mod;
	cout<<(f[x]=ans)<<endl;
}

signed main()
{
	n=read();
	for (int i=1;i<=n;i++)sc[i]=c[i]=read(),sc[i]+=sc[i-1],xx=xx*(c[i]+1)%Mod;
	for (int i=1;i<n;i++)d[i]=read();
	WT{solve(read());}
	return 0;
}
```