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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

std::string to_string(std::string s) { return '"' + s + '"'; }

std::string to_string(const char* s) { return to_string((std::string)s); }

std::string to_string(bool b) { return (b ? "true" : "false"); }

template <typename A, typename B>
std::string to_string(std::pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A>
std::string to_string(A v) {
  using ::to_string;
  using std::to_string;
  bool first = true;
  std::string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}

void debug_out() { std::cerr << std::endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  using ::to_string;
  using std::to_string;
  std::cerr << " " << to_string(H);
  debug_out(T...);
}

#ifdef LOCAL
#define debug(...)					\
  std::cerr << __FUNCTION__ << "(" << __LINE__ << ")"	\
  << "[" << #__VA_ARGS__ << "]:",			\
    debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

const int mod = 1e9 + 7;

inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }

inline int sub(int a, int b) { return a - b < 0 ? a - b + mod : a - b; }

inline int mul(int a, int b) { return (int)((long long)a * b % mod); }


int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n;
  cin >> n;
  vector<int> c(n);
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
  }
  vector<int> b(n);
  for (int i = 1; i < n; ++i) {
    cin >> b[i];
  }

  for (int i = 1; i < n; ++i) {
    b[i] += b[i - 1];
  }
  for (int i = 1; i < n; ++i) {
    b[i] += b[i - 1];
  }
  int sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += c[i];
  }

  int q;
  cin >> q;
  map<int, int> ans;
  while (q--) {
    int x;
    cin >> x;
    if (ans.find(x) != ans.end()) {
      cout << ans[x] << endl;
      continue;
    }
    int t_sum = 0;
    bool ok = true;
    for (int i = 0; i < n; ++i) {
      t_sum += c[i];
      if (t_sum < x * (i + 1) + b[i]) {
	ok = false;
	break;
      }
    }
    if (!ok) {
      cout << 0 << endl;
      continue;
    }
    ok = true;
    for (int i = 0; i < n; ++i) {
      if (x * (i + 1) + b[i] > 0) {
	ok = false;
	break;
      }
    }
    if (ok) {
      int p = 1;
      for (int i = 0; i < n; ++i) {
	p = mul(p, c[i] + 1);
      }
      cout << p << endl;
      continue;
    }
    vector<int> f[2];
    f[0].resize(sum + 1);
    f[1].resize(sum + 1);
    int now = 0;
    int pre = 1;
    int l = max(0, x);
    int r = c[0];
    for (int k = l; k <= r; ++k) {
      f[now][k] = 1;
    }
    for (int k = l + 1; k <= sum; ++k) {
      f[now][k] = add(f[now][k], f[now][k - 1]);
    }
    for (int i = 1; i < n; ++i) {
      now ^= 1;
      pre ^= 1;
      fill(f[now].begin(), f[now].end(), 0);
      int l = max(0, x * (i + 1) + b[i]);
      int r = sum;
      for (int k = l; k <= r; ++k) {
	int vl = k - c[i] <= 0 ? 0 : f[pre][k - c[i] - 1];
	int vr = f[pre][k];
	f[now][k] = sub(vr, vl);
      }
      for (int k = l + 1; k <= sum; ++k) {
	f[now][k] = add(f[now][k], f[now][k - 1]);
      }
    }
    ans[x] = f[now][sum];
    cout << ans[x] << endl;
  }
  return 0;
		 
}