#include <bits/stdc++.h>
using namespace std;
 
mt19937_64 rng(time(0));
const long double eps = 1e-15;
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const int mn=300+10;
const ll mod=1e9+7;

ll po(ll a,ll b=mod-2){
	ll ans=1;
	for(;b;b>>=1,a*=a,a%=mod)if(b&1)ans*=a,ans%=mod;
	return ans;
}

ll c[mn],b[mn];
ll ans[mn*2];
ll dp[2][10100];

int main() {
	cin.tie(0);
	cin.sync_with_stdio(0);
	int n;
	cin>>n;
	for(int i=0;i<n;i++)cin>>c[i];
	for(int i=1;i<n;i++)cin>>b[i],b[i]+=b[i-1];
	for(int i=1;i<n;i++)b[i]+=b[i-1];
	int ma=INT_MAX,mi=INT_MAX,sa=0,si=0;
	for(int i=0;i<n;i++){
		sa+=c[i];
		si+=0;
		ma=min(ll(ma),(sa-b[i])/(i+1)+1);
		mi=min(ll(mi),(si-b[i])/(i+1)-1);
	}
	//cerr<<mi<<" "<<ma<<endl;
	
	for(int m=mi;m<=ma;m++){
		memset(dp,0,sizeof(dp));
		dp[0][0]=1;
		for(int i=0;i<n;i++){
			ll lim = m*(i+1)+b[i];
			//cerr<<lim<<endl;
			for(int j=1;j<10100;j++)dp[0][j]+=dp[0][j-1],dp[0][j]%=mod;
			for(int j=0;j<10100;j++){
				dp[1][j]=(dp[0][j]-((j>=c[i]+1)?dp[0][j-c[i]-1]:0))%mod;
			}
			for(int j=0;j<min(lim,ll(10100));j++)dp[1][j]=0;
			memcpy(dp[0],dp[1],sizeof(dp[0]));
			memset(dp[1],0,sizeof(dp[1]));
		}
		for(int i=0;i<10100;i++)ans[m-mi]+=dp[0][i],ans[m-mi]%=mod;
		if(ans[m-mi]<0)ans[m-mi]+=mod;
	}
	int q;
	cin>>q;
	ll aa=1;
	for(int i=0;i<n;i++)aa*=c[i]+1,aa%=mod;
	for(int i=0;i<q;i++){
		ll x;
		cin>>x;
		if(x<mi){
			printf("%lld\n",aa);
		}
		else if(x>ma)printf("0\n");
		else printf("%lld\n",ans[x-mi]);
	}
}

### 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>
using namespace std;
 
mt19937_64 rng(time(0));
const long double eps = 1e-15;
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const int mn=300+10;
const ll mod=1e9+7;

ll po(ll a,ll b=mod-2){
	ll ans=1;
	for(;b;b>>=1,a*=a,a%=mod)if(b&1)ans*=a,ans%=mod;
	return ans;
}

ll c[mn],b[mn];
ll ans[mn*2];
ll dp[2][10100];

int main() {
	cin.tie(0);
	cin.sync_with_stdio(0);
	int n;
	cin>>n;
	for(int i=0;i<n;i++)cin>>c[i];
	for(int i=1;i<n;i++)cin>>b[i],b[i]+=b[i-1];
	for(int i=1;i<n;i++)b[i]+=b[i-1];
	int ma=INT_MAX,mi=INT_MAX,sa=0,si=0;
	for(int i=0;i<n;i++){
		sa+=c[i];
		si+=0;
		ma=min(ll(ma),(sa-b[i])/(i+1)+1);
		mi=min(ll(mi),(si-b[i])/(i+1)-1);
	}
	//cerr<<mi<<" "<<ma<<endl;
	
	for(int m=mi;m<=ma;m++){
		memset(dp,0,sizeof(dp));
		dp[0][0]=1;
		for(int i=0;i<n;i++){
			ll lim = m*(i+1)+b[i];
			//cerr<<lim<<endl;
			for(int j=1;j<10100;j++)dp[0][j]+=dp[0][j-1],dp[0][j]%=mod;
			for(int j=0;j<10100;j++){
				dp[1][j]=(dp[0][j]-((j>=c[i]+1)?dp[0][j-c[i]-1]:0))%mod;
			}
			for(int j=0;j<min(lim,ll(10100));j++)dp[1][j]=0;
			memcpy(dp[0],dp[1],sizeof(dp[0]));
			memset(dp[1],0,sizeof(dp[1]));
		}
		for(int i=0;i<10100;i++)ans[m-mi]+=dp[0][i],ans[m-mi]%=mod;
		if(ans[m-mi]<0)ans[m-mi]+=mod;
	}
	int q;
	cin>>q;
	ll aa=1;
	for(int i=0;i<n;i++)aa*=c[i]+1,aa%=mod;
	for(int i=0;i<q;i++){
		ll x;
		cin>>x;
		if(x<mi){
			printf("%lld\n",aa);
		}
		else if(x>ma)printf("0\n");
		else printf("%lld\n",ans[x-mi]);
	}
}

```