#include<cstdio>
#include<map>
#include<algorithm>
#define mod (1000000007)
using namespace std;
int n,k,b[105],lc[105],rc[105],Sum,SumL,SumR;

int dp[105][10005],pf[105][10005],tag[105];
map<int,int> mp;
int getPF(int x,int y)
{
    if(y<tag[x]) return 0;
    if(y>10000+tag[x]) return pf[x][10000];
    return pf[x][y-tag[x]];
}
int DP(int x)
{
//    printf("DP:x=%d ---------------------------",x);
//    for(int i=1;i<=n;i++) printf("L=%d R=%d\n",lc[i]-x,rc[i]-x);
    if(mp[x]) return mp[x]-mod;
    for(int i=0;i<=n;i++) for(int j=0;j<=10000+1;j++) dp[i][j]=pf[i][j]=0;
    for(int i=0;i<=n;i++) tag[i]=0;
    dp[0][0]=1;
    for(int i=0;i<=10000;i++) pf[0][i]=1;
    for(int i=1;i<=n;i++)
    {
        tag[i]=tag[i-1]+lc[i]-x;
        for(int j=max(tag[i],0);j<=tag[i]+10000;j++)
        {
            dp[i][j-tag[i]]=(getPF(i-1,j-lc[i]+x)-getPF(i-1,j-rc[i]+x-1)+mod)%mod;
//            printf("dp[%d][%d]=%d\n",i,j,dp[i][j]);
        }
        pf[i][0]=dp[i][0];
        for(int j=1;j<=10000;j++)
        {
            pf[i][j]=(pf[i][j-1]+dp[i][j])%mod;
        }
    }
    mp[x]=pf[n][10000]+mod;
    return pf[n][10000];
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&rc[i]);
    }
    for(int i=1;i<=n-1;i++)
    {
        scanf("%d",&b[i]);
        Sum+=b[i];
        rc[i+1]-=Sum;
        lc[i+1]-=Sum;
    }
    for(int i=1;i<=n;i++)
    {
        SumL+=lc[i];
        SumR+=rc[i];
    }
    int Minn=SumL/n-1,Maxn=SumR/n+1;
    scanf("%d",&k);
    for(int i=1;i<=k;i++)
    {
        int x;
        scanf("%d",&x);
        x=min(x,Maxn);
        x=max(x,Minn);
        printf("%d\n",DP(x));
    }
}
### Prompt

Generate a CPP solution to the following 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.

### Response

```cpp
#include<cstdio>
#include<map>
#include<algorithm>
#define mod (1000000007)
using namespace std;
int n,k,b[105],lc[105],rc[105],Sum,SumL,SumR;

int dp[105][10005],pf[105][10005],tag[105];
map<int,int> mp;
int getPF(int x,int y)
{
    if(y<tag[x]) return 0;
    if(y>10000+tag[x]) return pf[x][10000];
    return pf[x][y-tag[x]];
}
int DP(int x)
{
//    printf("DP:x=%d ---------------------------",x);
//    for(int i=1;i<=n;i++) printf("L=%d R=%d\n",lc[i]-x,rc[i]-x);
    if(mp[x]) return mp[x]-mod;
    for(int i=0;i<=n;i++) for(int j=0;j<=10000+1;j++) dp[i][j]=pf[i][j]=0;
    for(int i=0;i<=n;i++) tag[i]=0;
    dp[0][0]=1;
    for(int i=0;i<=10000;i++) pf[0][i]=1;
    for(int i=1;i<=n;i++)
    {
        tag[i]=tag[i-1]+lc[i]-x;
        for(int j=max(tag[i],0);j<=tag[i]+10000;j++)
        {
            dp[i][j-tag[i]]=(getPF(i-1,j-lc[i]+x)-getPF(i-1,j-rc[i]+x-1)+mod)%mod;
//            printf("dp[%d][%d]=%d\n",i,j,dp[i][j]);
        }
        pf[i][0]=dp[i][0];
        for(int j=1;j<=10000;j++)
        {
            pf[i][j]=(pf[i][j-1]+dp[i][j])%mod;
        }
    }
    mp[x]=pf[n][10000]+mod;
    return pf[n][10000];
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&rc[i]);
    }
    for(int i=1;i<=n-1;i++)
    {
        scanf("%d",&b[i]);
        Sum+=b[i];
        rc[i+1]-=Sum;
        lc[i+1]-=Sum;
    }
    for(int i=1;i<=n;i++)
    {
        SumL+=lc[i];
        SumR+=rc[i];
    }
    int Minn=SumL/n-1,Maxn=SumR/n+1;
    scanf("%d",&k);
    for(int i=1;i<=k;i++)
    {
        int x;
        scanf("%d",&x);
        x=min(x,Maxn);
        x=max(x,Minn);
        printf("%d\n",DP(x));
    }
}
```