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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod=1e9+7;
int mul(int x,int y)
{
    return x*y%mod;
}
int add(int x,int y)
{
    x+=y;
    if(x>=mod)
    x-=mod;
    return x;
}
int sub(int x,int y)
{
    x-=y;
    if(x<0)
    x+=mod;
    return x;
}
void solve()
{
    int n;
    cin>>n;
    vector<int> c(n);
    for(auto& x:c)
    cin>>x;
    vector<int> b(n-1);
    for(auto& x:b)
    cin>>x;
    int q;
    cin>>q;
    vector<int> x(q);
    for(auto& xx:x)
    cin>>xx;
    int p=1;
    for(int i=0;i<n;i++)
    p=mul(p,c[i]+1);
    const int nax=1e5;
    int val=-nax;
    vector<int> bound(n);
    for(int i=0;i<n;i++)
    {
        bound[i]=mul(val,i+1);
        for(int k=0;k<i;k++)
        {
            bound[i]+=mul(b[k],i-k);
            bound[i]%=mod;
        }
    }
    vector<int> res(2*nax+1,0);
    while(val<=nax)
    {
        bool f=0,cc=1;
        for(int i=0;i<n;i++)
        {
            if(bound[i]>100*(i+1))
            {
                f=1;
                break;
            }
            if(bound[i]>0)
            cc=0;
        }
        if(f)
        break;
        if(cc)
        res[val+nax]=p;
        else
        {
            vector<int> dp(1,1);
            for(int i=0;i<n;i++)
            {
                vector<int> new_dp((int)dp.size()+c[i]+1,0);
                for(int j=0;j<dp.size();j++)
                {
                    new_dp[j]+=dp[j];
                    new_dp[j]%=mod;
                    new_dp[j+c[i]+1]=sub(new_dp[j+c[i]+1],dp[j]);
                }
                for(int j=0;j<new_dp.size()-1;j++)
                {
                    new_dp[j+1]+=new_dp[j];
                    new_dp[j+1]%=mod;
                }
                new_dp.pop_back();
                swap(dp,new_dp);
                for(int j=0;j<bound[i] and j<dp.size();j++)
                dp[j]=0;
            }
            res[val+nax]=accumulate(dp.begin(),dp.end(),0LL)%mod;
        }
        for(int i=0;i<n;i++)
        {
            bound[i]+=i+1;
            bound[i]%=mod;
        }
        val++;
    }
    for(int xx=0;xx<q;xx++)
    cout<<res[x[xx]+nax]<<'\n';
}
int32_t main()
{
    int t=1;
    //cin>>t;
    while(t--)
    solve();
    return 0;
}