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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__);fflush(stderr);
#else
	#define eprintf(...) 42
#endif

using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) {
	return (ull)rng() % B;
}

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

clock_t startTime;
double getCurrentTime() {
	return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}


const ll MOD = (ll)1e9 + 7;
ll add(ll x, ll y) {
	x += y;
	if (x >= MOD) return x - MOD;
	return x;
}
ll sub(ll x, ll y) {
	x -= y;
	if (x < 0) return x + MOD;
	return x;
}
ll mult(ll x, ll y) {
	return (x * y) % MOD;
}
ll bin_pow(ll x, ll p) {
	if (p == 0) return 1;
	if (p & 1) return mult(x, bin_pow(x, p - 1));
	return bin_pow(mult(x, x), p / 2);
}
ll rev(ll x) {
	return bin_pow(x, MOD - 2);
}


const int N = 102;
const int C = 10101;
int n;
int c[N];
int b[N];
int sumB[N];
int dp[2][N][C];
int ANS[N];
int m;

int getAvg(int x, int y) {
	if (x >= 0) return x / y;
	return (x - y + 1) / y;
}

int main()
{
	startTime = clock();
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%d", &c[i]);
	for (int i = 1; i < n; i++) {
		int x;
		scanf("%d", &x);
		b[i] = b[i - 1] - x;
		sumB[i] = sumB[i - 1] + b[i];
	}
	m = 0;
	int sum = 0;
	for (int i = 0; i < n; i++) {
		sum += b[i];
		m = min(m, getAvg(sum, i + 1));
	}
	//eprintf("m = %d\n", m);
	dp[0][N - 1][0] = 1;
	for (int it = 0; it < n; it++) {
		for (int i = 0; i < N; i++)
			for (int j = 0; j < C; j++)
				dp[1][i][j] = 0;
		for (int avg = 0; avg < N; avg++)
			for (int sm = 0; sm < C; sm++) {
				if (dp[0][avg][sm] == 0) continue;
				int w = dp[0][avg][sm];
				int r = c[it];
				int curSum = sm + sumB[it];
				while(r >= 0) {
					int curAvg = getAvg(curSum + r, it + 1);
					assert(curAvg >= m);
					curAvg = min(avg, curAvg - m);
					int l = (m + curAvg) * (it + 1) - curSum;
					l = max(l, 0);
					assert(l <= r);
					dp[1][curAvg][sm + l] = add(dp[1][curAvg][sm + l], w);
					dp[1][curAvg][sm + r + 1] = sub(dp[1][curAvg][sm + r + 1], w);
					r = l - 1;
				}
			}
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < C; j++) {
				if (j == 0)
					dp[0][i][j] = 0;
				else
					dp[0][i][j] = dp[0][i][j - 1];
				dp[0][i][j] = add(dp[0][i][j], dp[1][i][j]);
			}
		}
	}
	for (int i = 0; i < N; i++)
		for (int j = 0; j < C; j++)
			ANS[i] = add(ANS[i], dp[0][i][j]);
	for (int i = N - 2; i >= 0; i--)
		ANS[i] = add(ANS[i], ANS[i + 1]);
	int q;
	scanf("%d", &q);
	while(q--) {
		int x;
		scanf("%d", &x);
		x = min(x, m + N - 1);
		x = max(x, m);
		printf("%d\n", ANS[x - m]);
	}

	return 0;
}