#include <bits/stdc++.h>

const int P = 1000000007;

int main() {
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(0);
	
	int n;
	std::cin >> n;
	std::vector<int> c(n);
	for (int i = 0; i < n; ++i) {
		std::cin >> c[i];
	}
	std::vector<int> b(n);
	for (int i = 0; i < n - 1; ++i) {
		int v;
		std::cin >> v;
		b[i + 1] = b[i] + v;
	}

	const int B = 100000;
	std::vector<int> ans(2 * B + 1, -1);

	auto DP = [&](int x) -> int {
		if (ans[x + B] != -1) {
			return ans[x + B];
		}
		int sum = 0, lim = 0;
		bool all0 = true, flag = true;
		for (int i = 0; i < n; ++i) {
			sum += c[i], lim += x + b[i];
			all0 &= lim <= 0;
			flag &= lim <= sum;
		}
		if (all0 && ans[0] != -1) {
			return ans[x + B] = ans[0];
		}
		if (!flag) {
			return ans[x + B] = 0;
		}
		std::vector<int> f(sum + 1, 0);
		sum = 0, lim = 0;
		f[0] = 1;
		for (int i = 0; i < n; ++i) {
			sum += c[i], lim += x + b[i];
			int s = 0;
			for (int k = 0; k <= c[i]; ++k) {
				s = (s + f[sum - k]) % P;
			}
			for (int j = sum; j >= 0; --j) {
				int t = j < lim ? 0 : s;
				s = (s + P - f[j]) % P;
				if (j > c[i]) {
					s = (s + f[j - c[i] - 1]) % P;
				}
				f[j] = t;
			}
		}
		return ans[x + B] = std::accumulate(f.begin(), f.end(), 0ll) % P;
	};

	DP(-100000);

	int q;
	std::cin >> q;
	while (q--) {
		int x;
		std::cin >> x;
		std::cout << DP(x) << "\n";
	}
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>

const int P = 1000000007;

int main() {
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(0);
	
	int n;
	std::cin >> n;
	std::vector<int> c(n);
	for (int i = 0; i < n; ++i) {
		std::cin >> c[i];
	}
	std::vector<int> b(n);
	for (int i = 0; i < n - 1; ++i) {
		int v;
		std::cin >> v;
		b[i + 1] = b[i] + v;
	}

	const int B = 100000;
	std::vector<int> ans(2 * B + 1, -1);

	auto DP = [&](int x) -> int {
		if (ans[x + B] != -1) {
			return ans[x + B];
		}
		int sum = 0, lim = 0;
		bool all0 = true, flag = true;
		for (int i = 0; i < n; ++i) {
			sum += c[i], lim += x + b[i];
			all0 &= lim <= 0;
			flag &= lim <= sum;
		}
		if (all0 && ans[0] != -1) {
			return ans[x + B] = ans[0];
		}
		if (!flag) {
			return ans[x + B] = 0;
		}
		std::vector<int> f(sum + 1, 0);
		sum = 0, lim = 0;
		f[0] = 1;
		for (int i = 0; i < n; ++i) {
			sum += c[i], lim += x + b[i];
			int s = 0;
			for (int k = 0; k <= c[i]; ++k) {
				s = (s + f[sum - k]) % P;
			}
			for (int j = sum; j >= 0; --j) {
				int t = j < lim ? 0 : s;
				s = (s + P - f[j]) % P;
				if (j > c[i]) {
					s = (s + f[j - c[i] - 1]) % P;
				}
				f[j] = t;
			}
		}
		return ans[x + B] = std::accumulate(f.begin(), f.end(), 0ll) % P;
	};

	DP(-100000);

	int q;
	std::cin >> q;
	while (q--) {
		int x;
		std::cin >> x;
		std::cout << DP(x) << "\n";
	}
}

```