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> 
using namespace std;
namespace Ruri{
    #define ms(a,b) memset(a,b,sizeof(a))
    #define repi(i,a,b) for(int i=a,bbb=b;i<=bbb;++i)//attention reg int or reg ll ?
    #define repd(i,a,b) for(int i=a,bbb=b;i>=bbb;--i)
    #define reps(s) for(int i=head[s],v=e[i].to;i;i=e[i].nxt,v=e[i].to)
    #define ce(i,r) i==r?'\n':' '
    #define pb push_back 
    #define all(x) x.begin(),x.end()
    #define gmn(a,b) a=min(a,b)
    #define gmx(a,b) a=max(a,b)
    #define fi first
    #define se second                         
    typedef long long ll;
    typedef unsigned long long ull;
    typedef double db;
    const int infi=1e9;//infi较大，注意涉及inf相加时爆int
    const ll infl=4e18;
    inline ll ceil_div(ll a,ll b){ return (a+b-1)/b; }
    inline ll pos_mod(ll a,ll b){ return (a%b+b)%b; }
    //std::mt19937 rnd(time(0));//std::mt19937_64 rnd(time(0));
}
using namespace Ruri;
namespace Read{
    #define ss(a) scanf("%s",a)
    inline int ri(){ int x; scanf("%d",&x); return x; }
    inline ll rl(){ ll x;  scanf("%lld",&x); return x; }
    inline db rd(){ db x; scanf("%lf",&x); return x; }
}
namespace DeBug{
    #define pr(x) cout<<#x<<": "<<(x)<<endl
    #define pra(x,a,b) cout<<#x<<": "<<endl; \
            repi(i,a,b) cout<<x[i]<<" ";     \
            puts("");   
    #define FR(a) freopen(a,"r",stdin)
    #define FO(a) freopen(a,"w",stdout)
}
using namespace Read;
using namespace DeBug;
const int MAX_N=105;
const int mod=1e9+7;
inline void Add(int &x,int y){ x+=y;if(x>=mod) x-=mod; }
int n,b[MAX_N],c[MAX_N];
int sb[MAX_N];
int f[MAX_N*MAX_N],g[MAX_N*MAX_N];
map<int,int> ans;
int Solve(int x)
{
	repi(i,0,10001) g[i]=1;
	repi(i,1,n){
		int lim=i*x+sb[i];
		ms(f,0);
		for(int j=max(0,lim);j<=10001;++j) f[j]=(g[j]-(j-c[i]-1>=0?g[j-c[i]-1]:0)+mod)%mod;
		ms(g,0),g[0]=f[0];
		for(int j=1;j<=10001;++j) g[j]=(1ll*g[j-1]+f[j])%mod;
	}
	return g[10001];
}
int main()
{
	n=ri();
	int prod=1;
	repi(i,1,n) c[i]=ri(),prod=1ll*prod*(c[i]+1)%mod;
	repi(i,1,n-1) b[i]=ri();
	repi(i,1,n)repi(j,1,i-1) sb[i]+=(i-j)*b[j];
	int l=0,r=infi,mx_c=*max_element(c+1,c+n+1);
	repi(i,1,n) gmn(l,-((sb[i]-1)/i+1)),gmn(r,(n*mx_c-sb[i]-1)/i+1);
	repi(i,l,r) ans[i]=Solve(i);
	int Q=ri();
	while(Q--)
	{
		int x=ri();
		if(x>r) puts("0");
		else if(x<l) printf("%d\n",prod);
		else printf("%d\n",ans[x]);
	}
    return 0;
}
