#include<bits/stdc++.h>
#define MAXN 105
#define MAXM 10005
#define MAXQ 100005
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int n,q,a[MAXN],b[MAXN],c[MAXN];
int x[MAXQ];
int sb[MAXN],sa[MAXN];
int dp[MAXN][MAXM];
int ans[405];
void add(int &a,int b) {a+=b; if(a>=MOD) a-=MOD;}
void dec(int &a,int b) {a-=b; if(a<0) a+=MOD;}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&c[i]);
    for(int i=1;i<=n-1;i++) scanf("%d",&b[i]);
    sb[1]=0;
    for(int i=2;i<=n;i++) sb[i]=sb[i-1]+b[i-1];
    for(int i=2;i<=n;i++) sb[i]+=sb[i-1];
    int lb=INF,rb=INF;
    scanf("%d",&q);
    for(int i=1;i<=q;i++) scanf("%d",&x[i]);
    //calculating lower bound
    for(int i=1;i<=n;i++)
    {
        sa[i]=0;
        lb=min(lb,(sa[i]-sb[i])/i);
        while(lb*i>sa[i]-sb[i]) lb--;
    }
    //calculating upper bound
    for(int i=1;i<=n;i++)
    {
        sa[i]=sa[i-1]+c[i];
        rb=min(rb,(sa[i]-sb[i])/i);
        while(rb*i>sa[i]-sb[i]) rb--;
    }
    lb--; rb++;
    ll s=0;
    for(int val=lb;val<=rb;val++)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0]=1; dp[0][1]=MOD-1;
        for(int i=0;i<=n;i++)
        {
            for(int j=1;j<=10000;j++) add(dp[i][j],dp[i][j-1]);
            for(int j=0;j<=10000;j++)
            {
                if(i>0&&j-sb[i]<1LL*val*i) dp[i][j]=0; 
                if(!dp[i][j]) continue;
                if(i<n)
                {
                    add(dp[i+1][j],dp[i][j]);
                    dec(dp[i+1][j+c[i+1]+1],dp[i][j]);
                }
            }
        }
        for(int j=0;j<=10000;j++) add(ans[val-lb],dp[n][j]);
    }
    for(int i=1;i<=q;i++)
    {
        if(x[i]<lb) printf("%d\n",ans[0]);
        else if(x[i]>rb) printf("%d\n",0);
        else printf("%d\n",ans[x[i]-lb]);
    }
    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 MAXN 105
#define MAXM 10005
#define MAXQ 100005
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int n,q,a[MAXN],b[MAXN],c[MAXN];
int x[MAXQ];
int sb[MAXN],sa[MAXN];
int dp[MAXN][MAXM];
int ans[405];
void add(int &a,int b) {a+=b; if(a>=MOD) a-=MOD;}
void dec(int &a,int b) {a-=b; if(a<0) a+=MOD;}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&c[i]);
    for(int i=1;i<=n-1;i++) scanf("%d",&b[i]);
    sb[1]=0;
    for(int i=2;i<=n;i++) sb[i]=sb[i-1]+b[i-1];
    for(int i=2;i<=n;i++) sb[i]+=sb[i-1];
    int lb=INF,rb=INF;
    scanf("%d",&q);
    for(int i=1;i<=q;i++) scanf("%d",&x[i]);
    //calculating lower bound
    for(int i=1;i<=n;i++)
    {
        sa[i]=0;
        lb=min(lb,(sa[i]-sb[i])/i);
        while(lb*i>sa[i]-sb[i]) lb--;
    }
    //calculating upper bound
    for(int i=1;i<=n;i++)
    {
        sa[i]=sa[i-1]+c[i];
        rb=min(rb,(sa[i]-sb[i])/i);
        while(rb*i>sa[i]-sb[i]) rb--;
    }
    lb--; rb++;
    ll s=0;
    for(int val=lb;val<=rb;val++)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0]=1; dp[0][1]=MOD-1;
        for(int i=0;i<=n;i++)
        {
            for(int j=1;j<=10000;j++) add(dp[i][j],dp[i][j-1]);
            for(int j=0;j<=10000;j++)
            {
                if(i>0&&j-sb[i]<1LL*val*i) dp[i][j]=0; 
                if(!dp[i][j]) continue;
                if(i<n)
                {
                    add(dp[i+1][j],dp[i][j]);
                    dec(dp[i+1][j+c[i+1]+1],dp[i][j]);
                }
            }
        }
        for(int j=0;j<=10000;j++) add(ans[val-lb],dp[n][j]);
    }
    for(int i=1;i<=q;i++)
    {
        if(x[i]<lb) printf("%d\n",ans[0]);
        else if(x[i]>rb) printf("%d\n",0);
        else printf("%d\n",ans[x[i]-lb]);
    }
    return 0;
}
```