#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
typedef pair<ll,ll> pl;
typedef double db;
#define ls(x) ((x)<<1)
#define rs(x) ((x)<<1|1)
#define low(x) ((x)&-(x))
#define all(x) x.begin(),x.end()
#define mp make_pair
#define X first
#define Y second
#ifdef _DEBUG
    const int N=1e2+10;
#else
    const int N=1e2+10;
#endif
const ll mod=1e9+7;
//const ll mod=998244353;
inline ll gcd(ll a,ll b){return !b?a:gcd(b,a%b);}
inline ll q_pow(ll a,ll x){ll ans=1,tmp=a;while(x){if(x&1)(ans*=tmp)%=mod;(tmp*=tmp)%=mod;x>>=1;}return ans;}
template<typename T> inline void re(T &N){int f=1;char c;while((c=getchar())< '0'||c> '9')if(c=='-')f=-1;N=c-'0';while((c=getchar())>='0'&&c<='9')N=N*10+c-'0';N*=f;}
int m,n,t=1,st,en;
int c[N],b[N];
ll dp[N][N*N];
map<int,ll>ans;
inline ll val(int i,int l,int r)
{
    if(l<=0)return dp[i][r];
    else return (dp[i][r]-dp[i][l-1]+mod)%mod;
}
int main()
{
    #ifdef _DEBUG
        freopen("data.txt","r",stdin);
    #endif
    re(n);
    for(int i=1;i<=n;i++)re(c[i]);
    for(int i=2;i<=n;i++)re(b[i]);
    for(int i=1;i<=n;i++)b[i]+=b[i-1];
    for(int i=1;i<=n;i++)b[i]+=b[i-1];
    int M=*max_element(c+1,c+n+1);
    int l,r;
    r=M-(b[n]/n+(b[n]%n> 0));
    l=r-M;r++;
    for(int m=l;m<=r;m++)
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<=n*M;i++)dp[0][i]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=n*M;j++)
                if(j>=i*m+b[i])dp[i][j]=val(i-1,j-c[i],j);
            for(int j=1;j<=n*M;j++)
                (dp[i][j]+=dp[i][j-1])%=mod;
        }
        ans[m]=dp[n][n*M];
    }
    re(m);
    while(m--)
    {
        int x;re(x);
        x=max(x,l);
        x=min(x,r);
        printf("%lld\n",ans[x]);
    }
}

### 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;
typedef long long ll;
typedef pair<int,int> pi;
typedef pair<ll,ll> pl;
typedef double db;
#define ls(x) ((x)<<1)
#define rs(x) ((x)<<1|1)
#define low(x) ((x)&-(x))
#define all(x) x.begin(),x.end()
#define mp make_pair
#define X first
#define Y second
#ifdef _DEBUG
    const int N=1e2+10;
#else
    const int N=1e2+10;
#endif
const ll mod=1e9+7;
//const ll mod=998244353;
inline ll gcd(ll a,ll b){return !b?a:gcd(b,a%b);}
inline ll q_pow(ll a,ll x){ll ans=1,tmp=a;while(x){if(x&1)(ans*=tmp)%=mod;(tmp*=tmp)%=mod;x>>=1;}return ans;}
template<typename T> inline void re(T &N){int f=1;char c;while((c=getchar())< '0'||c> '9')if(c=='-')f=-1;N=c-'0';while((c=getchar())>='0'&&c<='9')N=N*10+c-'0';N*=f;}
int m,n,t=1,st,en;
int c[N],b[N];
ll dp[N][N*N];
map<int,ll>ans;
inline ll val(int i,int l,int r)
{
    if(l<=0)return dp[i][r];
    else return (dp[i][r]-dp[i][l-1]+mod)%mod;
}
int main()
{
    #ifdef _DEBUG
        freopen("data.txt","r",stdin);
    #endif
    re(n);
    for(int i=1;i<=n;i++)re(c[i]);
    for(int i=2;i<=n;i++)re(b[i]);
    for(int i=1;i<=n;i++)b[i]+=b[i-1];
    for(int i=1;i<=n;i++)b[i]+=b[i-1];
    int M=*max_element(c+1,c+n+1);
    int l,r;
    r=M-(b[n]/n+(b[n]%n> 0));
    l=r-M;r++;
    for(int m=l;m<=r;m++)
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<=n*M;i++)dp[0][i]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=n*M;j++)
                if(j>=i*m+b[i])dp[i][j]=val(i-1,j-c[i],j);
            for(int j=1;j<=n*M;j++)
                (dp[i][j]+=dp[i][j-1])%=mod;
        }
        ans[m]=dp[n][n*M];
    }
    re(m);
    while(m--)
    {
        int x;re(x);
        x=max(x,l);
        x=min(x,r);
        printf("%lld\n",ans[x]);
    }
}

```