#include <bits/stdc++.h>
using namespace std;
const int BASE = 1e9 + 7;
const int A = 100;

int n, c[111], b[111], q, sumB[111], f[111][10100], ans[111];

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

  int lower = -(sumB[n - 2] / n);
  for (int x0 = lower; x0 <= lower + A; x0++)
  {
    memset(f, 0, sizeof f);
    for (int i = max(0, x0); i <= c[0]; i++)
      f[0][i] = 1;

    for (int i = 1; i < n; i++)
    {
      for (int sum = 0; sum <= i * A; sum++)
        if (f[i - 1][sum])
        {
          int minNewSum = max(sum, x0 * (i + 1) + sumB[i - 1]);
          int maxNewSum = sum + c[i];
          if (minNewSum <= maxNewSum)
          {
            f[i][minNewSum] = (f[i][minNewSum] + f[i - 1][sum]) % BASE;
            f[i][maxNewSum + 1] = (f[i][maxNewSum + 1] - f[i - 1][sum] + BASE) % BASE;
          }
        }

      for (int sum = 1; sum <= n * A; sum++)
        f[i][sum] = (f[i][sum] + f[i][sum - 1]) % BASE;
    }

    for (int i = 0; i <= n * A; i++)
      ans[x0 - lower] = (ans[x0 - lower] + f[n - 1][i]) % BASE;
  }

  long long allWays = accumulate(c, c + n, 1LL, [](long long u, int v) { return u * (v + 1) % BASE; });

  cin >> q;
  while (q--)
  {
    int x;
    cin >> x;
    if (x > lower + A) cout << 0 << '\n';
    else if (x < lower) cout << allWays << '\n';
    else cout << ans[x - lower] << '\n';
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 BASE = 1e9 + 7;
const int A = 100;

int n, c[111], b[111], q, sumB[111], f[111][10100], ans[111];

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

  int lower = -(sumB[n - 2] / n);
  for (int x0 = lower; x0 <= lower + A; x0++)
  {
    memset(f, 0, sizeof f);
    for (int i = max(0, x0); i <= c[0]; i++)
      f[0][i] = 1;

    for (int i = 1; i < n; i++)
    {
      for (int sum = 0; sum <= i * A; sum++)
        if (f[i - 1][sum])
        {
          int minNewSum = max(sum, x0 * (i + 1) + sumB[i - 1]);
          int maxNewSum = sum + c[i];
          if (minNewSum <= maxNewSum)
          {
            f[i][minNewSum] = (f[i][minNewSum] + f[i - 1][sum]) % BASE;
            f[i][maxNewSum + 1] = (f[i][maxNewSum + 1] - f[i - 1][sum] + BASE) % BASE;
          }
        }

      for (int sum = 1; sum <= n * A; sum++)
        f[i][sum] = (f[i][sum] + f[i][sum - 1]) % BASE;
    }

    for (int i = 0; i <= n * A; i++)
      ans[x0 - lower] = (ans[x0 - lower] + f[n - 1][i]) % BASE;
  }

  long long allWays = accumulate(c, c + n, 1LL, [](long long u, int v) { return u * (v + 1) % BASE; });

  cin >> q;
  while (q--)
  {
    int x;
    cin >> x;
    if (x > lower + A) cout << 0 << '\n';
    else if (x < lower) cout << allWays << '\n';
    else cout << ans[x - lower] << '\n';
  }
}

```