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.

Solution (in CPP):

Solution:
/*
Author: QAQAutoMaton
Lang: C++
Code: C1.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define debug(qaq...) fprintf(stderr,qaq)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
#define unq(a) sort(all(a)),a.erase(unique(all(a)),a.end())
using namespace std;
typedef unsigned uint;
typedef long long ll;
typedef unsigned long long ull;
typedef complex<double> cp;
typedef pair<int,int> pii;
int inf;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T,class T2>int chkmin(T &a,T2 b){return a>b?a=b,1:0;}
template<class T,class T2>int chkmax(T &a,T2 b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T,class T2>T mmin(T a,T2 b){return a<b?a:b;}
template<class T,class T2>T mmax(T a,T2 b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
template<class T>bool sort2(T &a,T &b){return a>b?swap(a,b),1:0;}
#define min mmin
#define max mmax
#define abs aabs
struct __INIT__{
	__INIT__(){
		fill((unsigned char*)&inf,(unsigned char*)&inf+sizeof(inf),0x3f);
	}
}__INIT___;
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	template<typename A>
	inline bool read (A &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		while((x=gc())==' '||x=='\n' || x=='\r');
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r');
		if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	template<typename A>
	inline bool write (A x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
const int p=1000000007;
struct Z{
	uint x;
	Z(){}
	Z(uint a){
		x=a;
	}
};
inline uint modp(const uint x){
	return x<p?x:x-p;
}
inline Z operator+(const Z x1, const Z x2) { return modp(x1.x+x2.x);}
inline Z operator-(const Z x1, const Z x2) { return modp(x1.x+p-x2.x);}
inline Z operator-(const Z x) {return x.x?p-x.x:0;}
inline Z operator*(const Z x1, const Z x2) { return static_cast<ull>(x1.x)*x2.x%p;}
void exgcd(int a,int b,int &x,int &y){
	if(!b){x=1;y=0;return;}
	exgcd(b,a%b,y,x);
	y-=(a/b)*x;
}
inline Z Inv(const Z a){
	int x,y;
	exgcd(p,a.x,x,y);
	return y<0?y+=p:y;
}
inline Z operator/(const Z x1, const Z x2) { return x1*Inv(x2);}

inline Z &operator++(Z &x1){x1.x==p-1?x1.x=0:++x1.x;return x1;} 
inline Z &operator--(Z &x1){x1.x?--x1.x:x1.x=p-1;return x1;}
inline Z &operator+=(Z &x1, const Z x2) { return x1 = x1 + x2; }
inline Z &operator-=(Z &x1, const Z x2) { return x1 = x1 - x2; }
inline Z &operator*=(Z &x1, const Z x2) { return x1 = x1 * x2; }
inline Z &operator/=(Z &x1, const Z x2) { return x1 = x1 / x2; }
inline Z fpm(Z a,int b){Z c(1);for(;b;b>>=1,a*=a)if(b&1)c*=a;return c;}
int c[105],b[105];
Z f[10005],g[10005];
int w[105];
signed main(){
#ifdef QAQAutoMaton 
	freopen("C1.in","r",stdin);
	freopen("C1.out","w",stdout);
#endif
	int n;	
	read(n);
	int l=0;
	for(int i=1;i<=n;++i){
		read(c[i]);
	}

	for(int i=l;~i;--i)f[i]+=f[i+1];
	for(int i=1;i<n;++i){
		read(b[i]);

	}
	int ss=0;
	for(int i=2;i<=n;++i){
		ss+=b[i-1];
		w[i]=w[i-1]+ss;

	}
	int q;
	read(q);
	int x;
	for(;q;--q){
		read(x);	
		for(int i=0;i<=l;++i)f[i]=0;
		f[0]=1;
		l=0;
		for(int i=1;i<=n;++i){
			g[0]=f[0];
			l+=c[i];
			for(int j=1;j<=l;++j)g[j]=f[j]+g[j-1];
			for(int j=0;j<=l;++j){
				f[j]=g[j]-(j>c[i]?g[j-c[i]-1]:0);
			}
			for(int j=0;j<=l;++j){
				if(x*i+w[i]>j)f[j]=0;
			}
		}
		Z ans(0);
		for(int i=0;i<=l;++i)ans+=f[i];
		write(ans.x,'\n');
	}
	return 0;
}