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.
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

const int MAXN=110;
const int mod=1e9+7;
int b[MAXN],c[MAXN],bp[MAXN],bpp[MAXN],cp[MAXN],f[2][2*MAXN*MAXN],ans[MAXN<<1];

inline void add(int &x,int y)
{
	x+=y;
	if(x>=mod)
		x-=mod;
}

int main()
{
	int n,i,j,k;
	int mn,mx,x,q,nxt,cur;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%d",&c[i]);
		cp[i]=cp[i-1]+c[i];
	}
	for(i=1;i<n;i++)
	{
		scanf("%d",&b[i]);
		bp[i]=bp[i-1]+b[i];
		bpp[i]=bpp[i-1]+bp[i];
	}
	mn=1<<30;
	for(i=1;i<=n;i++)
		mn=min(mn,-bpp[i-1]/i);
	mx=mn+150;
	mn=mn-45;
	memset(ans,-1,sizeof(ans));
	scanf("%d",&q);
	while(q--)
	{
		scanf("%d",&x);
		x=min(mx,max(mn,x));
		if(~ans[x-mn])
		{
			printf("%d\n",ans[x-mn]);
			continue;
		}
		cur=1,nxt=0;
		memset(f,0,sizeof(f));
		for(j=0;j<=c[1];j++)
			f[cur][j]=1;
		for(i=1;i<=n;i++)
		{
			memset(f[nxt],0,sizeof(f[nxt]));
			for(j=max(x*i+bpp[i-1],0);j<=cp[i];j++)
			{
				add(f[nxt][j],f[cur][min(cp[i-1],j)]);
				if(j-c[i]-1>=0)
					add(f[nxt][j],mod-f[cur][j-c[i]-1]);
			}
			for(j=1;j<=cp[i];j++)
				add(f[nxt][j],f[nxt][j-1]);
			swap(cur,nxt);
		}
		ans[x-mn]=f[cur][cp[n]];
		printf("%d\n",ans[x-mn]);
	}
}
