#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n, c[111], b[111], q, x, dp[111][11111], pre[111][11111], sum[111], all, ans[333], l, r, mnl, mnr, cc;
void add(int &x, int y)
{
	x += y;
	if (x >= mod) x -= mod; 
}
int solve(int x)
{
	memset(dp, 0, sizeof(dp));
	dp[0][0] = 1;
	for (int i = 0; i <= all; i++) pre[0][i] = 1;
	for (int i = 1; i <= n; i++)
	{
		for (int j = sum[i - 1] + i * x; j <= all; j++)
		{
			dp[i][j] = pre[i - 1][j];
			if (c[i] < j) dp[i][j] = (dp[i][j] - pre[i - 1][j - c[i] - 1]) % mod;
		}
		pre[i][0] = dp[i][0];
		for (int j = 1; j <= all; j++) pre[i][j] = (pre[i][j - 1] + dp[i][j]) % mod; 
	}
	int res = 0;
	for (int i = 0; i <= all; i++) res = (res + dp[n][i]) % mod;
	return res;
}
int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
	for (int i = 1; i < n; i++) scanf("%d", &b[i]);
	cc = 1;
	for (int i = 1; i <= n; i++) 
	{
		cc = 1ll * cc * (c[i] + 1) % mod;
	}
	for (int i = 1; i < n; i++)
	{
		sum[i] = sum[i - 1] + b[i];
	}
	for (int i = 1; i <= n; i++) all += c[i];
	mnl = mnr = 1e9;
	for (int i = 1; i <= n; i++)
	{
		l -= sum[i - 1];
		r += (c[i] - sum[i - 1]);
		mnl = min(mnl, l / i);
		mnr = min(mnr, r / i);
	}
	for (int i = 1; i <= n; i++) sum[i] += sum[i - 1];
	for (int i = mnl; i <= mnr; i++) ans[i - mnl] = solve(i);
	scanf("%d", &q);
	while(q--)
	{
		scanf("%d", &x);
		if (x > mnr) printf("0\n");
		else if (x < mnl) printf("%d\n", cc);
		else printf("%d\n", (ans[x - mnl] + mod) % mod);
	}
	return 0;
}
### Prompt

Create a solution in cpp for 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 mod = 1e9 + 7;
int n, c[111], b[111], q, x, dp[111][11111], pre[111][11111], sum[111], all, ans[333], l, r, mnl, mnr, cc;
void add(int &x, int y)
{
	x += y;
	if (x >= mod) x -= mod; 
}
int solve(int x)
{
	memset(dp, 0, sizeof(dp));
	dp[0][0] = 1;
	for (int i = 0; i <= all; i++) pre[0][i] = 1;
	for (int i = 1; i <= n; i++)
	{
		for (int j = sum[i - 1] + i * x; j <= all; j++)
		{
			dp[i][j] = pre[i - 1][j];
			if (c[i] < j) dp[i][j] = (dp[i][j] - pre[i - 1][j - c[i] - 1]) % mod;
		}
		pre[i][0] = dp[i][0];
		for (int j = 1; j <= all; j++) pre[i][j] = (pre[i][j - 1] + dp[i][j]) % mod; 
	}
	int res = 0;
	for (int i = 0; i <= all; i++) res = (res + dp[n][i]) % mod;
	return res;
}
int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
	for (int i = 1; i < n; i++) scanf("%d", &b[i]);
	cc = 1;
	for (int i = 1; i <= n; i++) 
	{
		cc = 1ll * cc * (c[i] + 1) % mod;
	}
	for (int i = 1; i < n; i++)
	{
		sum[i] = sum[i - 1] + b[i];
	}
	for (int i = 1; i <= n; i++) all += c[i];
	mnl = mnr = 1e9;
	for (int i = 1; i <= n; i++)
	{
		l -= sum[i - 1];
		r += (c[i] - sum[i - 1]);
		mnl = min(mnl, l / i);
		mnr = min(mnr, r / i);
	}
	for (int i = 1; i <= n; i++) sum[i] += sum[i - 1];
	for (int i = mnl; i <= mnr; i++) ans[i - mnl] = solve(i);
	scanf("%d", &q);
	while(q--)
	{
		scanf("%d", &x);
		if (x > mnr) printf("0\n");
		else if (x < mnl) printf("%d\n", cc);
		else printf("%d\n", (ans[x - mnl] + mod) % mod);
	}
	return 0;
}
```