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:
#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif //_MSC_VER

#include <cstdio>

#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <numeric>
#include <ctime>
#include <chrono>
#include <iostream>
#include <string>
#include <sstream>
#include <array>
#include <random>
#include <cmath>
#include <cassert>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <utility>
#include <thread>

using std::cin;
using std::cout;
using std::string;
using std::vector;

using ll = long long;

constexpr int MOD = 1000000000 + 7;  
constexpr int N = 117;
int b[N], c[N], f[2][2 * N * N], ans[N * 2];

static int solve(int testcase)
{
	int n;
	cin >> n;
	for (int i = 1; i <= n; ++i)cin >> c[i];
	for (int i = 1; i <= n - 1; ++i)cin >> b[i];

	int mn = INT_MAX;
	for (int i = 1; i <= n; ++i) {
		int sum = 0;
		for (int j = 0; j < i; ++j) {
			for (int k = 1; k <= j; ++k) {
				sum += b[k];
			}

		}

		mn = std::min(mn, -sum/i);
	}

	int mx = mn + 150;
	mn = mn - 45;

	memset(ans, -1, sizeof(ans));

	int q;
	cin >> q;
	for (int iq = 1; iq <= q; ++iq)
	{
		int x;
		cin >> x;
		x = std::min(mx, std::max(x, mn));

		if (ans[x - mn] != -1)
		{
			cout << ans[x - mn] << '\n';
			continue;
		}

		int res = 0;
		int delta = 0, sum = 0, sb = 0;
		int cur = 1, nxt = 0;
		memset(f, 0, sizeof(f));
		f[cur][0] = 1;

		for (int i = 1; i <= n; ++i) {
			memset(f[nxt], 0, sizeof(f[nxt]));
			delta += sb;
			for (int j = 0; j <= sum; ++j) {
				int presum = x * i + delta - j;
				int bb = std::max(presum, 0);

				if (bb <= c[i]) {
					f[nxt][j + bb] += f[cur][j];
					f[nxt][j + c[i] + 1] += MOD - f[cur][j];
					if (f[nxt][j + bb] >= MOD)
						f[nxt][j + bb] -= MOD;
					if (f[nxt][j + c[i] + 1] >= MOD)
						f[nxt][j + c[i] + 1] -= MOD;
				}
			}

			sum += c[i];

			for (int j = 1; j <= sum; ++j) {
				f[nxt][j] += f[nxt][j - 1];
				if (f[nxt][j] >= MOD)
					f[nxt][j] -= MOD;
			}

			sb += b[i];
			std::swap(cur, nxt);
		}
		for (int i = 0; i <= sum; ++i)
		{
			res += f[cur][i];
			if (res >= MOD)
				res -= MOD;
		}
		ans[x - mn] = res;
		cout << res << '\n';

	}

	return 0;
}

int main(int argc, char* argv[])
{
 
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif //ONLINE_JUDGE
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);
	
	int tc = 1;
	//std::cin >> tc;
	for (int ic = 1; ic <= tc; ++ic)
	{
		solve(ic);
	}

	return 0;
}