#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 105, MOD = 1e9 + 7;
int n, f[N * N], g[N * N], b[N], sb[N], c[N];
map<int, int> ans;
int Solve(int x)
{
	fill(g, g + N * N, 1);
	for(int i = 1, sumc = c[1], lim; i <= n; i++, sumc += c[i]) 
	{
		lim = i * x + sb[i]; memset(f, 0, sizeof(f));
		for(int j = max(0, lim); j < N * N; j++) f[j] = (g[j] - (j - c[i] - 1 >= 0 ? g[j - c[i] - 1] : 0) + MOD) % MOD;
		memset(g, 0, sizeof(g)); g[0] = f[0];
		for(int j = 1; j < N * N; j++) g[j] = (g[j - 1] + f[j]) % MOD;
	}
	return g[N * N - 1];
}
int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	int prod = 1;
	for(int i = 1; i <= n; i++) { cin >> c[i]; prod = (ll)prod * (c[i] + 1) % MOD; }
	for(int i = 1; i < n; i++) cin >> b[i];
	for(int i = 1; i <= n; i++) for(int j = 1; j < i; j++) sb[i] += (i - j) * b[j];
	int lb = 0, rb = INT_MAX, m = *max_element(c + 1, c + n + 1);
	for(int i = 1; i <= n; i++) lb = min(lb, -((sb[i] - 1) / i + 1)), rb = min(rb, (n * m - sb[i] - 1) / i + 1);
	for(int i = lb; i <= rb; i++) ans[i] = Solve(i);
	int q; cin >> q;
	while(q--)
	{
		int x; cin >> x;
		if(x >= lb && x <= rb) cout << ans[x] << endl;
		else if(x > rb) cout << 0 << endl;
		else if(x < lb) cout << prod << endl;
	}
	return 0;
}
### Prompt

Please create a solution in cpp 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;
typedef long long ll;
const int N = 105, MOD = 1e9 + 7;
int n, f[N * N], g[N * N], b[N], sb[N], c[N];
map<int, int> ans;
int Solve(int x)
{
	fill(g, g + N * N, 1);
	for(int i = 1, sumc = c[1], lim; i <= n; i++, sumc += c[i]) 
	{
		lim = i * x + sb[i]; memset(f, 0, sizeof(f));
		for(int j = max(0, lim); j < N * N; j++) f[j] = (g[j] - (j - c[i] - 1 >= 0 ? g[j - c[i] - 1] : 0) + MOD) % MOD;
		memset(g, 0, sizeof(g)); g[0] = f[0];
		for(int j = 1; j < N * N; j++) g[j] = (g[j - 1] + f[j]) % MOD;
	}
	return g[N * N - 1];
}
int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	int prod = 1;
	for(int i = 1; i <= n; i++) { cin >> c[i]; prod = (ll)prod * (c[i] + 1) % MOD; }
	for(int i = 1; i < n; i++) cin >> b[i];
	for(int i = 1; i <= n; i++) for(int j = 1; j < i; j++) sb[i] += (i - j) * b[j];
	int lb = 0, rb = INT_MAX, m = *max_element(c + 1, c + n + 1);
	for(int i = 1; i <= n; i++) lb = min(lb, -((sb[i] - 1) / i + 1)), rb = min(rb, (n * m - sb[i] - 1) / i + 1);
	for(int i = lb; i <= rb; i++) ans[i] = Solve(i);
	int q; cin >> q;
	while(q--)
	{
		int x; cin >> x;
		if(x >= lb && x <= rb) cout << ans[x] << endl;
		else if(x > rb) cout << 0 << endl;
		else if(x < lb) cout << prod << endl;
	}
	return 0;
}
```