#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
using VI = vector<int>;
using VL = vector<LL>;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define MP make_pair
#define PB push_back
#define EB emplace_back
#define F first
#define S second
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define FILL(a, b) memset(a, b, sizeof(a))

void dout() { cerr << endl; }

template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
    cerr << H << ' ';
    dout(T...);
}
const int mod = 1e9 + 7;
int add(int a, int b){
	if(a + b < mod) return a + b;
	return a + b - mod;
}
int sub(int a, int b){
	if(a < b) return a - b + mod;
	return a - b;
}
int mult(int a, int b){
	return a * (LL) b % mod;
}
const int M = 107;
int ans[2 * M];
int dp[M][M * M + 2 * M];
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n;
	cin >> n;
	VI c(n), b(n - 1);
	FOR(i, 0, n) cin >> c[i];
	FOR(i, 0, n - 1) cin >> b[i];
	FOR(i, 0, 2) FOR(j, 0, n - 1) b[j] = b[j] + (j != 0 ? b[j - 1] : 0);
	int l = -(b[n - 2] / n) - 10;
	FOR(x, l, l + 2 * M)
	{
		FILL(dp, 0);
		dp[0][0] = 1;
		dp[0][1] = -1;
		FOR(i, 0, n)
		FOR(sum, 0, M * M)
		{
			int mn_ai = max(0, (i + 1) * x + (i > 0 ? b[i - 1] : 0) - sum);
			if(sum)
				dp[i][sum] = add(dp[i][sum], dp[i][sum - 1]);
			if(mn_ai > c[i]) continue;
			dp[i + 1][sum + mn_ai] = add(dp[i + 1][sum + mn_ai], dp[i][sum]);
			dp[i + 1][sum + c[i] + 1] = sub(dp[i + 1][sum + c[i] + 1], dp[i][sum]);
		}
		int res = 0;
		FOR(sum, 0, M * M)
		{
			if(sum) dp[n][sum] = add(dp[n][sum], dp[n][sum - 1]);
			res = add(res, dp[n][sum]);
		}
		ans[x - l] = res;
	}
	int tmp = 1;
	FOR(i, 0, n)
		tmp = mult(tmp, c[i] + 1);
	int q;
	cin >> q;
	while(q--){
		int x;
		cin >> x;
		x = max(x, l);
		x = min(x, l + 2 * M - 1);
		cout << ans[x - l] << endl;
	}
	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;
using LL = long long;
using ULL = unsigned long long;
using VI = vector<int>;
using VL = vector<LL>;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define MP make_pair
#define PB push_back
#define EB emplace_back
#define F first
#define S second
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define FILL(a, b) memset(a, b, sizeof(a))

void dout() { cerr << endl; }

template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
    cerr << H << ' ';
    dout(T...);
}
const int mod = 1e9 + 7;
int add(int a, int b){
	if(a + b < mod) return a + b;
	return a + b - mod;
}
int sub(int a, int b){
	if(a < b) return a - b + mod;
	return a - b;
}
int mult(int a, int b){
	return a * (LL) b % mod;
}
const int M = 107;
int ans[2 * M];
int dp[M][M * M + 2 * M];
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n;
	cin >> n;
	VI c(n), b(n - 1);
	FOR(i, 0, n) cin >> c[i];
	FOR(i, 0, n - 1) cin >> b[i];
	FOR(i, 0, 2) FOR(j, 0, n - 1) b[j] = b[j] + (j != 0 ? b[j - 1] : 0);
	int l = -(b[n - 2] / n) - 10;
	FOR(x, l, l + 2 * M)
	{
		FILL(dp, 0);
		dp[0][0] = 1;
		dp[0][1] = -1;
		FOR(i, 0, n)
		FOR(sum, 0, M * M)
		{
			int mn_ai = max(0, (i + 1) * x + (i > 0 ? b[i - 1] : 0) - sum);
			if(sum)
				dp[i][sum] = add(dp[i][sum], dp[i][sum - 1]);
			if(mn_ai > c[i]) continue;
			dp[i + 1][sum + mn_ai] = add(dp[i + 1][sum + mn_ai], dp[i][sum]);
			dp[i + 1][sum + c[i] + 1] = sub(dp[i + 1][sum + c[i] + 1], dp[i][sum]);
		}
		int res = 0;
		FOR(sum, 0, M * M)
		{
			if(sum) dp[n][sum] = add(dp[n][sum], dp[n][sum - 1]);
			res = add(res, dp[n][sum]);
		}
		ans[x - l] = res;
	}
	int tmp = 1;
	FOR(i, 0, n)
		tmp = mult(tmp, c[i] + 1);
	int q;
	cin >> q;
	while(q--){
		int x;
		cin >> x;
		x = max(x, l);
		x = min(x, l + 2 * M - 1);
		cout << ans[x - l] << endl;
	}
	return 0;
}

```