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 rep(i, a, b) for(int i = a; i < b; i ++)
#define mod 1000000007
#define LL long long
using namespace std;


long long f[2][10005], SUM[10005], res[100000];
int c[105], n, b[105],  sb[105], ssb[105], sa[105], lim[105];

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
#endif

	cin.tie(0); cout.tie(0);
	ios::sync_with_stdio(0);

	int ss = 0;
	
	cin >> n;
	for (int i = 0; i < n ; i ++) cin >> c[i];
	for (int i = 0; i < n - 1; i ++) cin >> b[i];
	sa[0] = c[0];
	for (int i = 0; i < n - 1; i ++) {
		if (!i) sb[i] = b[i];
		else sb[i] = sb[i - 1] + b[i];
		sa[i + 1] = sa[i] + c[i + 1];
	}
	int mn = 1000000000;
	for (int i = 0; i < n - 1; i ++) {
		if (!i) ssb[i] = sb[i];
		else ssb[i] = ssb[i - 1] + sb[i];
		mn = min(mn, -ssb[i] / (i + 1));
	}
	int mx = mn + 250;
	mn = mn - 10;
	int q, x;
	memset(res, -1, sizeof(res));
	cin >> q;
	while(q --) {
		cin >> x;
		x = max(mn, min(mx, x));
		if (res[x - mn] > -1) {
			cout << res[x - mn] << endl;
			continue;
		}
		memset(f, 0, sizeof(f));
		int xx = x;
		lim[0] = x;
		for (int i = 0; i < n - 1; i ++) x += b[i], lim[i + 1] = lim[i] + x;//, cerr << x << ' ';
		int id = 1;
		x = xx;
		for (int i = max(x, 0); i <= c[0]; i ++) {
			f[0][i] = i - max(x, 0) + 1;
		}
		for (int i = 1; i < n; i ++) {
			for (int ii = 0; ii < 10005; ii ++) f[id][ii] = 0;
			for (int j = max(0, lim[i]); j <= sa[i]; j ++) {
				int st = max(j - c[i] - 1, -1), en = min(j, sa[i - 1]);
				if (st > sa[i - 1]) continue;
				if (en < lim[i - 1]) continue;
				f[id][j] = f[1 - id][en];
				if (st > -1) f[id][j] = (f[id][j] - f[1 - id][st] + mod) % mod;
			}
			for (int j = max(0, lim[i]) + 1; j <= sa[i]; j ++) {				
				f[id][j] = (f[id][j] + f[id][j - 1]) % mod;
			}
			
			if (i < n - 1)id = 1 - id;
		}
		res[x - mn] = f[id][sa[n - 1]];
		cout << res[x - mn] << endl;
	} 
	return 0;
}