using namespace std;

#include <iostream>
#include <cstring>

#define MOD 1000000007
#define MAXN 105

int N,K,Q,a[MAXN],b[MAXN],f[MAXN],dp[MAXN * MAXN];

int main() {
	cin >> N;
	K = N * MAXN;
	for(int i = 1;i <= N;++i) {
		cin >> a[i];
	}
	for(int i = 2;i <= N;++i) {
		cin >> b[i];
		b[i] += b[i - 1];
	}
	int s = 1e9;
	int e = 1e9;
	int pre = 0;
	for(int i = 1;i <= N;++i) {
		pre += a[i];
		b[i] += b[i - 1];
		s = min(s,-b[i] / i - 1);
		e = min(e,(pre - b[i]) / i + 1);
	}
	for(int i = s;i <= e;++i) {
		memset(dp,0,sizeof(dp));
		dp[0] = 1;
		for(int j = 1;j <= N;++j) {
			for(int k = 1;k <= K;++k) {
				dp[k] += dp[k - 1];
				if(dp[k] >= MOD) dp[k] -= MOD;
			}
			for(int k = K;~k;--k) {
				if(k < i * j + b[j]) {
					dp[k] += MOD - dp[k];
				}else if(k > a[j]) {
					dp[k] += MOD - dp[k - a[j] - 1];
				}
				dp[k] %= MOD;
			}
		}
		for(int j = 0;j <= K;++j) {
			f[i - s] += dp[j];
			if(f[i - s] >= MOD) f[i - s] -= MOD;
		}
	}
	cin >> Q;
	while(Q--) {
		int x;
		cin >> x;
		x = max(s,min(x,e));
		cout << f[x - s] << endl;
	}
	return 0;
}
### Prompt

Please formulate 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
using namespace std;

#include <iostream>
#include <cstring>

#define MOD 1000000007
#define MAXN 105

int N,K,Q,a[MAXN],b[MAXN],f[MAXN],dp[MAXN * MAXN];

int main() {
	cin >> N;
	K = N * MAXN;
	for(int i = 1;i <= N;++i) {
		cin >> a[i];
	}
	for(int i = 2;i <= N;++i) {
		cin >> b[i];
		b[i] += b[i - 1];
	}
	int s = 1e9;
	int e = 1e9;
	int pre = 0;
	for(int i = 1;i <= N;++i) {
		pre += a[i];
		b[i] += b[i - 1];
		s = min(s,-b[i] / i - 1);
		e = min(e,(pre - b[i]) / i + 1);
	}
	for(int i = s;i <= e;++i) {
		memset(dp,0,sizeof(dp));
		dp[0] = 1;
		for(int j = 1;j <= N;++j) {
			for(int k = 1;k <= K;++k) {
				dp[k] += dp[k - 1];
				if(dp[k] >= MOD) dp[k] -= MOD;
			}
			for(int k = K;~k;--k) {
				if(k < i * j + b[j]) {
					dp[k] += MOD - dp[k];
				}else if(k > a[j]) {
					dp[k] += MOD - dp[k - a[j] - 1];
				}
				dp[k] %= MOD;
			}
		}
		for(int j = 0;j <= K;++j) {
			f[i - s] += dp[j];
			if(f[i - s] >= MOD) f[i - s] -= MOD;
		}
	}
	cin >> Q;
	while(Q--) {
		int x;
		cin >> x;
		x = max(s,min(x,e));
		cout << f[x - s] << endl;
	}
	return 0;
}
```