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>
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;

string _reset = "\u001b[0m", _yellow = "\u001b[33m", _bold = "\u001b[1m";
void DBG() { cerr << "]" << _reset << endl; }
template<class H, class...T> void DBG(H h, T ...t) {
	cerr << to_string(h);
	if(sizeof ...(t)) cerr << ", ";
	DBG(t...);
}
#ifdef CHEISSMART
#define debug(...) cerr << _yellow << _bold << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define debug(...)
#endif

const int INF = 1e9 + 7, N = 105, M = 1e9 + 7;

int c[N], b[N], at_least[N], aux[N], dp[N][N*N];

map<int, int> ans;

void add(int& a, int b) {
	a += b;
	if(a >= M) a -= M;
}

signed main()
{
	IO_OP;

	int n;
	cin >> n;
	int ways = 1;
	for(int i = 1; i <= n; i++) {
		cin >> c[i];
		ways = (ll) ways * (c[i] + 1) % M;
	}
	for(int i = 1; i <= n - 1; i++) {
		cin >> b[i];
	}
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= i; j++)
			for(int k = 1; k < j; k++)
				aux[i] += b[k];
	auto solve = [&] (int x) {
		for(int i = 1; i <= n; i++) {
			at_least[i] = aux[i] + i * x;
			at_least[i] = max(at_least[i], 0);
		}
		memset(dp, 0, sizeof dp);
		dp[0][0] = 1;
		for(int i = 0; i < n; i++) {
			auto add_range = [&] (int l, int r, int x) {
				if(l > r) return;
				add(dp[i + 1][l], x);
				add(dp[i + 1][r + 1], M - x);
			};		
			for(int j = at_least[i]; j < N * N; j++) if(dp[i][j]) {
				int need = max(at_least[i + 1] - j, 0);
				add_range(j + need, j + c[i + 1], dp[i][j]);
			}
			for(int j = 1; j < N * N; j++)
				add(dp[i + 1][j], dp[i + 1][j - 1]);
		}
		int ans = 0;
		for(int j = 0; j < N * N; j++)
			add(ans, dp[n][j]);
		return ans;
	};
	int lb = -(aux[n] + n - 1) / n, rb = lb + 100;
 	for(int i = lb; i <= rb; i++)
 		ans[i] = solve(i);
 	int q;
 	cin >> q;
 	while(q--) {
 		int x;
 		cin >> x;
 		if(x < lb)
 			cout << ways << '\n';
 		else if(x <= rb)
 			cout << ans[x] << '\n';
 		else 
 			cout << 0 << '\n';
 	}


}