#include<iostream>
#include<cstdio>
#include<unordered_map>
using namespace std;
const int N=105;
const int MOD=1000000007;
int n,q;
int c[N];
int b[N];
int sb[N],sc[N];
int l,r;
int solve(int x)
{
    static int f[N][N*N],g[N][N*N];
    f[0][0]=g[0][0]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<max(i*x+sb[i-1],0);j++)
            f[i][j]=0;
        for(int j=max(i*x+sb[i-1],0);j<=sc[i];j++)
            f[i][j]=((long long)g[i-1][j-max(j-sc[i-1],0)]-(j-min(c[i],j)-1>=0?g[i-1][j-min(c[i],j)-1]:0)+MOD)%MOD;
        g[i][0]=f[i][0];
        for(int j=1;j<=sc[i];j++)
            g[i][j]=(g[i][j-1]+f[i][j])%MOD;
    }
    int ans=0;
    for(int j=0;j<=sc[n];j++)
        ans=(ans+f[n][j])%MOD;
    return ans;
}
unordered_map<int,int>res;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&c[i]);
    for(int i=1;i<n;i++)
        scanf("%d",&b[i]);
    for(int i=1;i<n;i++)
        sb[i]=sb[i-1]+b[i];
    for(int i=1;i<n;i++)
        sb[i]+=sb[i-1];
    for(int i=1;i<=n;i++)
        sc[i]=sc[i-1]+c[i];
    r=(sc[n]-sb[n-1])/n;
    for(int i=1;i<n;i++)
        r=min(r,(sc[i]-sb[i-1])/i);
    l=-sb[n-1]/n;
    for(int i=1;i<n;i++)
        l=min(l,sb[i-1]/i);
    for(int i=l;i<=r;i++)
        res[i]=solve(i);
    int pw=1;
    for(int i=1;i<=n;i++)
        pw=1LL*pw*(c[i]+1)%MOD;
    scanf("%d",&q);
    while(q--)
    {
        int x;
        scanf("%d",&x);
        if(x<l) printf("%d\n",pw);
        else if(x>r) printf("%d\n",0);
        else printf("%d\n",res[x]);
    }
    return 0;
}
### Prompt

Generate a Cpp solution to the following problem:

This is the easy version of the problem. The only difference is that in this version q = 1. 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 (q=1).

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
1
-1


Output


56

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<iostream>
#include<cstdio>
#include<unordered_map>
using namespace std;
const int N=105;
const int MOD=1000000007;
int n,q;
int c[N];
int b[N];
int sb[N],sc[N];
int l,r;
int solve(int x)
{
    static int f[N][N*N],g[N][N*N];
    f[0][0]=g[0][0]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<max(i*x+sb[i-1],0);j++)
            f[i][j]=0;
        for(int j=max(i*x+sb[i-1],0);j<=sc[i];j++)
            f[i][j]=((long long)g[i-1][j-max(j-sc[i-1],0)]-(j-min(c[i],j)-1>=0?g[i-1][j-min(c[i],j)-1]:0)+MOD)%MOD;
        g[i][0]=f[i][0];
        for(int j=1;j<=sc[i];j++)
            g[i][j]=(g[i][j-1]+f[i][j])%MOD;
    }
    int ans=0;
    for(int j=0;j<=sc[n];j++)
        ans=(ans+f[n][j])%MOD;
    return ans;
}
unordered_map<int,int>res;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&c[i]);
    for(int i=1;i<n;i++)
        scanf("%d",&b[i]);
    for(int i=1;i<n;i++)
        sb[i]=sb[i-1]+b[i];
    for(int i=1;i<n;i++)
        sb[i]+=sb[i-1];
    for(int i=1;i<=n;i++)
        sc[i]=sc[i-1]+c[i];
    r=(sc[n]-sb[n-1])/n;
    for(int i=1;i<n;i++)
        r=min(r,(sc[i]-sb[i-1])/i);
    l=-sb[n-1]/n;
    for(int i=1;i<n;i++)
        l=min(l,sb[i-1]/i);
    for(int i=l;i<=r;i++)
        res[i]=solve(i);
    int pw=1;
    for(int i=1;i<=n;i++)
        pw=1LL*pw*(c[i]+1)%MOD;
    scanf("%d",&q);
    while(q--)
    {
        int x;
        scanf("%d",&x);
        if(x<l) printf("%d\n",pw);
        else if(x>r) printf("%d\n",0);
        else printf("%d\n",res[x]);
    }
    return 0;
}
```