#include<bits/stdc++.h>
using namespace std;
inline int readint(){
	int x=0;
	bool f=0;
	char c=getchar();
	while(!isdigit(c)&&c!='-') c=getchar();
	if(c=='-'){
		f=1;
		c=getchar();
	}
	while(isdigit(c)){
		x=x*10+c-'0';
		c=getchar();
	}
	return f?-x:x;
}
const int maxn=100+5;
int n,c[maxn],b[maxn],q;
int sb[maxn];
typedef long long ll;
const ll mod=1e9+7;
ll dp[maxn*2][2][maxn*maxn],s[maxn*maxn];
int main(){
	#ifdef LOCAL
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
	#endif
	n=readint();
	for(int i=1;i<=n;i++) c[i]=readint();
	for(int i=1;i<n;i++) b[i]=readint();
	for(int i=1;i<n;i++) sb[i]=sb[i-1]+b[i];
	for(int i=1;i<n;i++) sb[i]+=sb[i-1];
	ll prod=1;
	for(int i=1;i<=n;i++) prod=prod*(c[i]+1)%mod;
	int mn=-sb[n-1]/n-1,mx=2e9;
	for(int i=1;i<n;i++) mx=min(mx,(maxn*i-sb[i-1])/i+1);
	for(int i=mn;i<=mx;i++){
		for(int j=0;j<maxn*maxn;j++) dp[i-mn][(n+1)%2][j]=1;
		for(int j=n;j>0;j--){
			s[0]=dp[i-mn][(j+1)%2][0];
			for(int k=0;k<maxn*maxn;k++)
				s[k]=(s[k-1]+dp[i-mn][(j+1)%2][k])%mod;
			for(int k=0;k<maxn*maxn;k++){
				int l=max(k,i*j+sb[j-1]),r=min(k+c[j],maxn*maxn-1);
				if(l>r) dp[i-mn][j%2][k]=0;
				else dp[i-mn][j%2][k]=(s[r]-(l?s[l-1]:0)+mod)%mod;
			}
		}
	}
	int q=readint();
	while(q--){
		int x=readint();
		if(x<mn) printf("%lld\n",prod);
		else if(x>mx) printf("0\n");
		else printf("%lld\n",dp[x-mn][1][0]);
	}
	return 0;
}

### Prompt

Please create a solution in CPP 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;
inline int readint(){
	int x=0;
	bool f=0;
	char c=getchar();
	while(!isdigit(c)&&c!='-') c=getchar();
	if(c=='-'){
		f=1;
		c=getchar();
	}
	while(isdigit(c)){
		x=x*10+c-'0';
		c=getchar();
	}
	return f?-x:x;
}
const int maxn=100+5;
int n,c[maxn],b[maxn],q;
int sb[maxn];
typedef long long ll;
const ll mod=1e9+7;
ll dp[maxn*2][2][maxn*maxn],s[maxn*maxn];
int main(){
	#ifdef LOCAL
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
	#endif
	n=readint();
	for(int i=1;i<=n;i++) c[i]=readint();
	for(int i=1;i<n;i++) b[i]=readint();
	for(int i=1;i<n;i++) sb[i]=sb[i-1]+b[i];
	for(int i=1;i<n;i++) sb[i]+=sb[i-1];
	ll prod=1;
	for(int i=1;i<=n;i++) prod=prod*(c[i]+1)%mod;
	int mn=-sb[n-1]/n-1,mx=2e9;
	for(int i=1;i<n;i++) mx=min(mx,(maxn*i-sb[i-1])/i+1);
	for(int i=mn;i<=mx;i++){
		for(int j=0;j<maxn*maxn;j++) dp[i-mn][(n+1)%2][j]=1;
		for(int j=n;j>0;j--){
			s[0]=dp[i-mn][(j+1)%2][0];
			for(int k=0;k<maxn*maxn;k++)
				s[k]=(s[k-1]+dp[i-mn][(j+1)%2][k])%mod;
			for(int k=0;k<maxn*maxn;k++){
				int l=max(k,i*j+sb[j-1]),r=min(k+c[j],maxn*maxn-1);
				if(l>r) dp[i-mn][j%2][k]=0;
				else dp[i-mn][j%2][k]=(s[r]-(l?s[l-1]:0)+mod)%mod;
			}
		}
	}
	int q=readint();
	while(q--){
		int x=readint();
		if(x<mn) printf("%lld\n",prod);
		else if(x>mx) printf("0\n");
		else printf("%lld\n",dp[x-mn][1][0]);
	}
	return 0;
}

```