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;

// 5 is a root of both mods
template <int MOD, int RT> struct Mint {
	static const int mod = MOD;
	static constexpr Mint rt() { return RT; } // primitive root for FFT
	int v; 
	explicit operator int() const { return v; } // explicit -> don't silently convert to int
	Mint() { v = 0; }
	Mint(long long _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD); if (v < 0) v += MOD; }
	friend bool operator==(const Mint& a, const Mint& b) { return a.v == b.v; }
	friend bool operator!=(const Mint& a, const Mint& b) { return !(a == b); }
	friend bool operator<(const Mint& a, const Mint& b) { return a.v < b.v; }
	friend bool operator>(const Mint& a, const Mint& b) { return a.v > b.v; }
	friend bool operator<=(const Mint& a, const Mint& b) { return a.v <= b.v; }
	friend bool operator>=(const Mint& a, const Mint& b) { return a.v >= b.v; }
	friend std::istream& operator >> (std::istream& in, Mint& a) { 
		long long x; std::cin >> x; a = Mint(x); return in; }
	friend std::ostream& operator << (std::ostream& os, const Mint& a) { return os << a.v; }
	Mint& operator+=(const Mint& m) { 
		if ((v += m.v) >= MOD) v -= MOD; 
		return *this; }
	Mint& operator-=(const Mint& m) { 
		if ((v -= m.v) < 0) v += MOD; 
		return *this; }
	Mint& operator*=(const Mint& m) { 
		v = (long long) v * m.v % MOD; return *this; }
	Mint& operator/=(const Mint& m) { return (*this) *= inv(m); }
	friend Mint pow(Mint a, long long p) {
		Mint ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p & 1) ans *= a;
		return ans; 
	}
	friend Mint inv(const Mint& a) { assert(a.v != 0); return pow(a, MOD - 2); }
	Mint operator-() const { return Mint(-v); }
	Mint& operator++() { return *this += 1; }
	Mint& operator--() { return *this -= 1; }
	friend Mint operator+(Mint a, const Mint& b) { return a += b; }
	friend Mint operator-(Mint a, const Mint& b) { return a -= b; }
	friend Mint operator*(Mint a, const Mint& b) { return a *= b; }
	friend Mint operator/(Mint a, const Mint& b) { return a /= b; }
};

using mi = Mint<1000000007, 5>;

const int N = 105;

mi dp[N][N * N];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n;
	cin >> n;
	vector<int> c(n + 1);
	mi prod = 1;
	for (int i = 1; i <= n; ++i) {
		cin >> c[i];
		prod *= c[i] + 1;
	}
	vector<int> b(n);
	for (int i = 1; i <= n - 1; ++i) {
		cin >> b[i];
	}
	vector<int> p = b;
	for (int t = 0; t < 2; ++t) {
		for (int i = 1; i <= n - 1; ++i) {
			p[i] += p[i - 1];
		}
	}
	auto solve = [&](int x) {
		for (int i = 0; i < N; ++i) {
			for (int j = 0; j < N * N; ++j) {
				dp[i][j] = 0;
			}
		}
		dp[0][0] = 1;
		vector<mi> prefix(N * N);
		for (int i = 1; i <= n; ++i) {
			for (int j = 0; j < N * N; ++j) {
				prefix[j] = (j ? prefix[j - 1] : 0) + dp[i - 1][j];
			}
			for (int j = 0; j < N * N; ++j) {
				if (j < i * x + p[i - 1]) {
					continue;
				}
				int l = max(j - c[i], 0);
				int r = j;
				dp[i][j] = prefix[r] - (l ? prefix[l - 1] : 0);
			}
		}
		mi ans = 0;
		for (int i = 0; i < N * N; ++i) {
			ans += dp[n][i];
		}
		return ans;
	};
	map<int, mi> res;
	int q;
	cin >> q;
	while (q--) {
		int x;
		cin >> x;
		if (res.find(x) != res.end()) {
			cout << res[x] << '\n';
		} else {
			if (N * N < x * n + p[n - 1]) {
				res[x] = 0;
			} else {
				bool ok = true;
				for (int i = 1; i <= n; ++i) {
					if (i * x + p[i - 1] >= 0) {
						ok = false;
					}
				}
				if (ok) {
					res[x] = prod;
				} else {
					res[x] = solve(x);
				}
			}
			cout << res[x] << '\n';
		}
	}
	return 0;
}