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;

const int p=1e9+7;
int dp[110][10010]={0};
int n,b[110],c[110],q,ans[10200]={0},range=0;

int main() {
    cin>>n;
    for(int i=0;i<n;i++)cin>>c[i];
    for(int i=0;i<n-1;i++){cin>>b[i];range+=(n-i-1)*b[i];}
    range=-1-range/n;
    for(q=range-10;q<=range+110;q++)
    {
        memset(dp,0,sizeof(dp));
        int lmt=q,sumb=0,sumc=c[0];
        for(int i=max(0,q);i<=sumc+1;i++)dp[0][i]=i-max(0,q);
        for(int i=sumc+2;i<10010;i++)dp[0][i]=dp[0][sumc+1];
        for(int i=1;i<n;i++)
        {
            sumb+=b[i-1];
            lmt+=sumb+q;
            sumc+=c[i];
            for(int j=max(0,lmt);j<=sumc;j++)
            {
                dp[i][j]=(dp[i-1][j+1]-dp[i-1][j-min(c[i],j)]+p)%p;
            }
            for(int j=max(0,lmt);j<=sumc;j++)dp[i][sumc+1]=(dp[i][sumc+1]+dp[i][j])%p;
            for(int j=sumc;j>=max(0,lmt);j--)dp[i][j]=(dp[i][j+1]-dp[i][j]+p)%p;
            for(int j=sumc+2;j<10010;j++)dp[i][j]=dp[i][sumc+1];
        }
        ans[q+10000]=dp[n-1][sumc+1];
    }
    cin>>q;
    while(cin>>q)
    {
        if(q>range+110)cout<<0<<endl;
        else if(q<range-10)cout<<ans[range+9990]<<endl;
        else cout<<ans[q+10000]<<endl;
    }
    return 0;
}