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.
// Bhagya Kamal Jain

#include<bits/stdc++.h>

#define ll          long long
#define pb          push_back
#define endl        '\n'
#define mii         map<ll int,ll int>
#define pii         pair<ll int,ll int>
#define vi          vector<ll int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define hell        1000000007
#define sim         template < class c
#define ris         return * this
#define dor         > debug & operator <<
#define eni(x)      sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {

using namespace std;

sim > struct rge {c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cout << *x, 0);
sim > char dud(...);
struct debug 
{
	~debug() 
	{ 
		cout<<endl; 
	}
	eni(!=) cout << boolalpha << i; ris; }
	eni(==) ris << range(begin(i), end(i)); }
	sim, class b dor(pair < b, c > d) 
	{
		ris << "(" << d.first << ", " << d.second << ")";
	}
	sim dor(rge<c> d) 
	{
	  	*this << "[";
	  	for (auto it = d.b; it != d.e; ++it)
	    	*this << ", " + 2 * (it == d.b) << *it;
	  	ris << "]";
	}
};
#define fuck(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

#define N  10005

ll int dp[N];
ll int tmp[N];

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int TESTS=1;
	// cin>>TESTS;
	while(TESTS--)
	{
		ll int n;
		cin >> n;
		ll int c[n + 1];
		for(ll int i = 1; i <= n; i++) cin >> c[i];
		ll int b[n];
		for(ll int i = 1; i < n; i++) cin >> b[i];
		
		ll int q;
		cin >> q;
		while(q--)
		{
			ll int x;
			cin >> x;
			ll int arr[n + 1];
			arr[1] = x;
			for(ll int i = 2; i <= n; i++)
			{
				arr[i] = arr[i - 1] + b[i - 1];
			}
			ll int val[n + 1];
			val[0] = 0;
			for(ll int i = 1; i <= n; i++)
			{
				val[i] = val[i - 1] + arr[i];
			}

		dp[0] = 1;
		for(ll int i = 1; i <= n; i++)
		{
			for(ll int j = 0; j < N; j++)
			{
				tmp[j] = dp[j];
			}
			memset(dp, 0, sizeof(dp));
			for(ll int j = 0; j < N; j++)
			{
				for(ll int k = 0; k <= c[i]; k++)
				{
					if(k + j < N) dp[k + j] = (dp[k + j] + tmp[j]) % hell;
				}
			}
			for(ll int j = 0; j < N; j++)
			{
				if(j < val[i]) dp[j] = 0;
			}
		}

			ll int ans = 0;
			for(ll int j = 0; j < N; j++)
			{
				ans = (ans + dp[j]) % hell;
			}
			cout << ans << endl;
		}
	}
	return 0;
}