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 ll long long
#define ull unsigned long long
#define ld long double
#define pii pair <int, int>
#define pll pair <ll, ll>
#define pci pair <char, int>
#define pld pair <ld, ld>
#define ppld pair <pld, pld>
#define ppll pair <pll, pll>
#define pldl pair <ld, ll>
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define cd complex <double>
#define PI 3.14159265358979
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll INF = 1e18;
ll mod = 1e9 + 7;
ll b[105], c[105];
ll prefix_b[105], prefix_c[105];
ll dp[105][10005], prefix[105][10005];
ll ans[205];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	ll n; cin >> n;
	for (ll i = 1; i <= n; i++) cin >> c[i];
	for (ll i = 2; i <= n; i++) cin >> b[i];
	for (ll i = 1; i <= n; i++){
		prefix_b[i] = prefix_b[i - 1] + b[i];
		prefix_c[i] = prefix_c[i - 1] + c[i];
	}
	for (ll i = 1; i <= n; i++) prefix_b[i] = prefix_b[i - 1] + prefix_b[i];
	dp[0][0] = 1;
	ll mn = INF, mx = INF;
	for (ll i = 1; i <= n; i++){
		ll lw = -prefix_b[i] / i - (prefix_b[i] % i != 0);
		ll hi = (prefix_c[i] - prefix_b[i] + 1) / i;
		if (prefix_c[i] - prefix_b[i] + 1 < 0) hi -= ((prefix_c[i] - prefix_b[i] + 1) % i != 0);
		mx = min(mx, hi);
		mn = min(mn, lw);
	}
	for (ll x = mn; x <= mx; x++){
		for (ll p = 0; p <= prefix_c[n]; p++) prefix[0][p] = 1;
		for (ll i = 1; i <= n; i++){
			for (ll p = 0; p <= prefix_c[i]; p++) dp[i][p] = 0;
			for (ll p = 0; p <= prefix_c[n]; p++) prefix[i][p] = 0;
		}
		for (ll i = 1; i <= n; i++){
			for (ll p = 0; p <= prefix_c[i]; p++){
				if (p - prefix_b[i] < x * i) continue;
				dp[i][p] = prefix[i - 1][p];
				if (p - c[i] > 0){
					dp[i][p] -= prefix[i - 1][p - c[i] - 1];
					if (dp[i][p] < 0) dp[i][p] += mod;
				}
			}
			prefix[i][0] = dp[i][0];
			for (ll p = 1; p <= prefix_c[n]; p++){
				prefix[i][p] = prefix[i][p - 1] + dp[i][p];
				if (prefix[i][p] >= mod) prefix[i][p] -= mod;
			}
	 	}
	 	ans[x - mn] = prefix[n][prefix_c[n]];
	}
	ll q; cin >> q;
	ll all = 1;
	for (ll i = 1; i <= n; i++) (all *= (c[i] + 1)) %= mod;
	while (q--){
		ll x; cin >> x;
		if (x < mn) cout << all << "\n";
		else if (x > mx) cout << 0 << "\n";
		else cout << ans[x - mn] << "\n";
	}
}