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;

const int MOD = 1e9+7;

vector<int> solve(const vector<int>& c, const vector<int>& b, const vector<int>& x) {
  int n = c.size();
  int q = x.size();
  vector<int> cb(n);
  cb[0] = 0;
  for (int i = 0; i < n-1; ++i) {
    cb[i+1] = cb[i] + b[i];
  }
  vector<int> d(n);
  d[0] = 0;
  for (int i = 0; i < n-1; ++i) {
    d[i+1] = d[i] + cb[i+1];
  }
  int low = -1e8, high = 1e8;
  while (low < high) {
    int mid = high - (high - low) / 2;
    bool ok = true;
    for (int i = 0; ok && (i < n); ++i) {
      if ((d[i] + int64_t(i+1)*mid) > 0) {
	ok = false;
      }
    }
    if (ok) {
      low = mid;
    }
    else {
      high = mid-1;
    }
  }
  vector<int> cc(n+1);
  cc[0] = 0;
  for (int i = 0; i < n; ++i) {
    cc[i+1] = cc[i] + c[i];
  }
  int maxsum = cc[n];
  auto calc = [&](int xi) {
    vector<int> dp(maxsum+1, 1);
    vector<int> cs(maxsum+2);
    cs[0] = 0;
    for (int i = n-1; i >= 0; --i) {
      vector<int> dp2(maxsum+1, 0);
      for (int j = 0; j <= maxsum; ++j) {
	cs[j+1] = cs[j] + dp[j];
	if (cs[j+1] >= MOD) cs[j+1] -= MOD;
      }
      for (int j = 0; j <= cc[i]; ++j) {
	dp2[j] = cs[j+c[i]+1] + (MOD - cs[j+min(max(0, d[i]+(i+1)*xi - j), c[i]+1)]);
	if (dp2[j] >= MOD) dp2[j] -= MOD;
      }
      dp2.swap(dp);
    }
    return dp[0];
  };
  vector<int> r(1, calc(low));
  for (int i = low+1; r.back() > 0; ++i) {
    r.push_back(calc(i));
  }
  vector<int> ans(q);
  for (int qi = 0; qi < q; ++qi) {
    ans[qi] = r[min(max(x[qi]-low, 0), int(r.size())-1)];
  }
  return ans;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> c(n);
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
  }
  vector<int> b(n-1);
  for (int i = 0; i < n-1; ++i) {
    cin >> b[i];
  }
  int q;
  cin >> q;
  vector<int> x(q);
  for (int i = 0; i < q; ++i) {
    cin >> x[i];
  }
  vector<int> solush = solve(c, b, x);
  for (int i = 0; i < q; ++i) {
    if (i > 0) cout << " ";
    cout << solush[i];
  }
  cout << endl;
}