Problem: This is the hard version of the problem. The only difference is that in this version 1 ≤ q ≤ 10^5. 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 (1 ≤ q ≤ 10^5).

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
5
-1 0 1 -100000 100000


Output


56
28
4
60
0

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:
#include <bits/stdc++.h>
using namespace std;

typedef long long Int;
typedef pair<int,int> PII;
typedef vector<int> VInt;
 
#define FOR(i, a, b) for(i = (a); i < (b); ++i)
#define RFOR(i, a, b) for(i = (a) - 1; i >= (b); --i)
#define EACH(it, a) for(auto it = (a).begin(); it != (a).end(); ++it)
#define CLEAR(a, b) memset(a, b, sizeof(a))
#define SIZE(a) int((a).size())
#define ALL(a) (a).begin(),(a).end()
#define MP make_pair

#define MOD (int(1e9+7))
#define MAX (10 << 10)

int C[128];
int B[128];
int S[128];
int L[128];
int R[128];
Int A[2][MAX];
Int Z[MAX];

Int f(int n, int x)
{
	int i, j;
	int sr = 0;
	int sl = 0;
	Int all = 1;
	bool fail = false;
	FOR(i, 0, n) 
	{
		int a = x + S[i];
		L[i] = 0 - a;
		R[i] = C[i] - a;
		sl += L[i];
		sr += R[i];
		all = all*(R[i] - L[i] + 1) % MOD;
		if(sr < 0) return 0;
		if(sl < 0) fail = true;
	}

	if(!fail) return all;

	sl = max(0, L[0]);
	sr = R[0];
	int len = sr - sl + 1;
	FOR(i, 0, len) A[1][i] = 1;
	FOR(i, 1, n)
	{
		int prev = i & 1;
		int next = prev ^ 1;
		Z[0] = 0;
		FOR(j, 0, len) Z[j + 1] = Z[j] + A[prev][j];

		int ll = max(0, sl + L[i]);
		int rr = sr + R[i];
		int len2 = rr - ll + 1;
		FOR(j, 0, len2)
		{
			int from = max(0, ll + j - R[i] - sl);
			int to = min(len, ll + j - L[i] + 1 - sl);
			A[next][j] = (Z[to] - Z[from]) % MOD;
		}
		sl = ll;
		sr = rr;
		len = len2;
	}

	Int res = 0;
	FOR(i, 0, len) res = (res + A[n & 1][i]) % MOD;
	return res;
}

void SolveTest(int test)
{
	int n;
	cin >> n;
	int i;
	FOR(i, 0, n) 
		cin >> C[i];
	FOR(i, 0, n - 1)
		cin >> B[i];
	S[0] = 0;
	FOR(i, 1, n) S[i] = S[i - 1] + B[i - 1];
	int q;
	cin >> q;
	vector<PII> qq(q);
	FOR(i, 0, q) 
	{
		cin >> qq[i].first;
		qq[i].second = i;
	}

	vector<long> res(q);
	sort(ALL(qq));
	FOR(i, 0, q)
	{
		if(i != 0 && qq[i].first == qq[i - 1].first)
			res[qq[i].second] = res[qq[i - 1].second];
		else
			res[qq[i].second] = f(n, qq[i].first);
	}

	FOR(i, 0, q) cout << res[i] << endl;
}

int main()
{
	int T, t;
	T = 1;
//	cin >> T;
	FOR(t, 0, T) SolveTest(t);
	return 0;
}