#include<bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
typedef double db;
typedef vector<int> vi;
typedef pair<int,int>pi;
const int maxn=1e6+10;
const int M=1000000007;
int n,x,b[105],s[105],c[105],dp[105][10005],S[10005],sum,tot,q,ans[105],best,sumc[105];
void add(int &x,int y){x+=y;if (x>=M)x-=M;}
void sub(int &x,int y){x-=y;if (x<0) x+=M;}
int _round(int x,int y){while (x%y) x--;return x/y;}
int main(){
    scanf("%d",&n); tot=1;
    for (int i=1;i<=n;i++) {
        scanf("%d",&c[i]);
        tot=1ll*tot*(c[i]+1)%M;
        sumc[i]=sumc[i-1]+c[i];
    }
    for (int i=1;i<n;i++) scanf("%d",&b[i]);
    for (int i=1;i<n;i++) b[i]+=b[i-1],s[i]=s[i-1]+b[i];
    best=c[1]; for (int i=2;i<=n;i++) best=min(best,_round(sumc[i]-s[i-1],i));
    for (int p=0;p<=101;p++){
        memset(dp,0,sizeof(dp));
        sum=0; dp[0][0]=1; x=best-p;
        for (int i=1;i<=n;i++){
            sum+=c[i]; S[0]=dp[i-1][0];
            for (int j=1;j<=sum;j++) S[j]=(S[j-1]+dp[i-1][j])%M;
            for (int j=0;j<=sum;j++){
                if (j<x*i+s[i-1]) dp[i][j]=0;
                else {
                    dp[i][j]=S[j];
                    if (j>c[i]) sub(dp[i][j],S[j-c[i]-1]);
                }
            }
        }
        ans[p]=0; for (int i=0;i<=sum;i++) add(ans[p],dp[n][i]);
    }
    scanf("%d",&q);
    while (q--){
        int x; scanf("%d",&x);
        if (x>best) puts("0"); else printf("%d\n",ans[min(101,best-x)]);
    }
    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
#define pb push_back
using namespace std;
typedef long long ll;
typedef double db;
typedef vector<int> vi;
typedef pair<int,int>pi;
const int maxn=1e6+10;
const int M=1000000007;
int n,x,b[105],s[105],c[105],dp[105][10005],S[10005],sum,tot,q,ans[105],best,sumc[105];
void add(int &x,int y){x+=y;if (x>=M)x-=M;}
void sub(int &x,int y){x-=y;if (x<0) x+=M;}
int _round(int x,int y){while (x%y) x--;return x/y;}
int main(){
    scanf("%d",&n); tot=1;
    for (int i=1;i<=n;i++) {
        scanf("%d",&c[i]);
        tot=1ll*tot*(c[i]+1)%M;
        sumc[i]=sumc[i-1]+c[i];
    }
    for (int i=1;i<n;i++) scanf("%d",&b[i]);
    for (int i=1;i<n;i++) b[i]+=b[i-1],s[i]=s[i-1]+b[i];
    best=c[1]; for (int i=2;i<=n;i++) best=min(best,_round(sumc[i]-s[i-1],i));
    for (int p=0;p<=101;p++){
        memset(dp,0,sizeof(dp));
        sum=0; dp[0][0]=1; x=best-p;
        for (int i=1;i<=n;i++){
            sum+=c[i]; S[0]=dp[i-1][0];
            for (int j=1;j<=sum;j++) S[j]=(S[j-1]+dp[i-1][j])%M;
            for (int j=0;j<=sum;j++){
                if (j<x*i+s[i-1]) dp[i][j]=0;
                else {
                    dp[i][j]=S[j];
                    if (j>c[i]) sub(dp[i][j],S[j-c[i]-1]);
                }
            }
        }
        ans[p]=0; for (int i=0;i<=sum;i++) add(ans[p],dp[n][i]);
    }
    scanf("%d",&q);
    while (q--){
        int x; scanf("%d",&x);
        if (x>best) puts("0"); else printf("%d\n",ans[min(101,best-x)]);
    }
    return 0;
}

```