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>

#define MAXN 111
#define BASE 1000000007

using namespace std;

int n;
int b[MAXN], c[MAXN];
int x, q;
int Dp[MAXN][MAXN * MAXN];
int prefixSumB[MAXN];
vector<pair<int, int>> preRes;

int DpCalc(int curX, int totalSum) {
  for (int i = 0; i <= n; i++) {
    for (int sum = 0; sum <= totalSum; sum++) {
      Dp[i][sum] = 0;
    }
  }
  Dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int sum = 0; sum <= totalSum; sum++) {
      int minA = i * curX + prefixSumB[i - 1] - sum;
      minA = max(minA, 0);
      if (minA <= c[i - 1]) {
        Dp[i][sum + minA] = (Dp[i][sum + minA] + Dp[i - 1][sum]) % BASE;
        Dp[i][sum + c[i - 1] + 1] = (Dp[i][sum + c[i - 1] + 1] - Dp[i - 1][sum] + BASE) % BASE;
      }
    }
    for (int sum = 0; sum <= totalSum; sum++) {
      if (sum > 0) Dp[i][sum] = (Dp[i][sum] + Dp[i][sum - 1]) % BASE;
    }
  }
  int res = 0;
  for (int sum = 0; sum <= totalSum; sum++) res = (res + Dp[n][sum]) % BASE;
  return res;
}

int main() {
  cin >> n;
  int total = 0;
  int maxC = 0;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    total += c[i];
    maxC = max(maxC, c[i]);
  } 
  b[0] = prefixSumB[0] = 0;
  int sumB = 0;
  for (int i = 1; i < n; i++) {
    cin >> b[i];
    sumB += b[i];
    prefixSumB[i] = prefixSumB[i - 1] + sumB;
  }
  // pre calc
  x = ((maxC + 1)* n - prefixSumB[n - 1]) / n;
  preRes.push_back(make_pair(x, DpCalc(x, total)));
  while (maxC >= 0) {
    x = (maxC * n - prefixSumB[n - 1]) / n;
    preRes.push_back(make_pair(x, DpCalc(x, total)));
    maxC--;
  }
  x = -100000;
  preRes.push_back(make_pair(x, DpCalc(x, total)));
  // end pre calc
  cin >> q;
  while (q--) {
    cin >> x;
    int res = 0;
    for (int i = 0; i < preRes.size(); i++) {
      if (x >= preRes[i].first) {
        res = preRes[i].second;
        break;
      }
    }
    cout << res << endl;
  }
  return 0;
}