#include <bits/stdc++.h>

using namespace std;

const int N = 105, mod = 1e9+7, inf = 1e9;
typedef long long ll;

int n, q;
int b[N], c[N], sb[N];
ll prod = 1, f[N*N], s[N*N];
map<int, ll> mp;

inline void solve(int x) {
	for (int i = 0; i <= n * 100; i++) s[i] = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j <= n * 100; j++) f[j] = 0;
		for (int j = max(0, i * x + sb[i]); j <= n * 100; j++)
			f[j] = (s[j] - (j - c[i] > 0 ? s[j-c[i]-1] : 0) + mod) % mod;
		s[0] = f[0];
		for (int j = 1; j <= n * 100; j++) s[j] = (s[j-1] + f[j]) % mod;
	}
	mp[x] = s[n*100];
}

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) 
		scanf("%d", &c[i]), prod = prod * (c[i] + 1) % mod;
	for (int i = 1; i < n; i++)
		scanf("%d", &b[i]);
	for (int i = 1; i <= n; i++) 
		for (int j = 1; j < i; j++)
			sb[i] += (i - j) * b[j];
	int L = inf, R = inf;
	for (int i = 2; i <= n; i++)
		L = min(L, -sb[i] / i), R = min(R, (100 * n - sb[i]) / i);
	for (int i = L; i <= R; i++) solve(i);
	scanf("%d", &q);
	for (int i = 1; i <= q; i++) {
		int x; scanf("%d", &x);
		if (x < L) printf("%lld\n", prod);
		else if (x > R) printf("0\n");
		else printf("%lld\n", mp[x]);
	}
	return 0;
}
### Prompt

Generate a CPP solution to the following 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.

### Response

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

using namespace std;

const int N = 105, mod = 1e9+7, inf = 1e9;
typedef long long ll;

int n, q;
int b[N], c[N], sb[N];
ll prod = 1, f[N*N], s[N*N];
map<int, ll> mp;

inline void solve(int x) {
	for (int i = 0; i <= n * 100; i++) s[i] = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j <= n * 100; j++) f[j] = 0;
		for (int j = max(0, i * x + sb[i]); j <= n * 100; j++)
			f[j] = (s[j] - (j - c[i] > 0 ? s[j-c[i]-1] : 0) + mod) % mod;
		s[0] = f[0];
		for (int j = 1; j <= n * 100; j++) s[j] = (s[j-1] + f[j]) % mod;
	}
	mp[x] = s[n*100];
}

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) 
		scanf("%d", &c[i]), prod = prod * (c[i] + 1) % mod;
	for (int i = 1; i < n; i++)
		scanf("%d", &b[i]);
	for (int i = 1; i <= n; i++) 
		for (int j = 1; j < i; j++)
			sb[i] += (i - j) * b[j];
	int L = inf, R = inf;
	for (int i = 2; i <= n; i++)
		L = min(L, -sb[i] / i), R = min(R, (100 * n - sb[i]) / i);
	for (int i = L; i <= R; i++) solve(i);
	scanf("%d", &q);
	for (int i = 1; i <= q; i++) {
		int x; scanf("%d", &x);
		if (x < L) printf("%lld\n", prod);
		else if (x > R) printf("0\n");
		else printf("%lld\n", mp[x]);
	}
	return 0;
}
```