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;

#define forn(I,N) for (int I = 0; I < N; ++I)
#define fornd(I,N) for (int I = N-1; I >= 0; --I)
#define forab(I,A,B) for (int I = A; I <= B; ++I)
#define forabd(I,A,B) for (int I = B; I >= A; --I)
#define foreach(I, A) for(typeof((A).begin()) I = (A).begin(); I != (A).end(); ++I)
// #define foreach(I, A) for((__typeof__(A))::iterator I = (A).begin(); I != (A).end(); ++I)

#define pb push_back
#define mp make_pair

typedef long long LL;

// sorting
struct P {
  int x, y;
  bool operator<(const P &p) {
    if (x != p.x) return x < p.x;
    else return y < p.y;
  }
};

/*
bool comp(data dt1, data dt2) {
  return dt1 < dt2;
}
*/

const LL mod = 1e9 + 7;

int main() {
  int n;
  cin >> n;

  vector<int> c(n), b(n);
  forn (i, n) {
    cin >> c[i];
  }
  forn (i, n - 1) {
    cin >> b[i + 1];
  }

  /*
  LL s1 = 0, s2 = 0;
  vector<map<LL, LL> > ct;
  map<LL, LL> bmap;
  bmap[1e6] = 1;
  ct.pb(bmap);
  forn (i, n) {
    cout << i << endl;
    s1 = (s1 + b[i]) % mod;
    s2 = (s2 + s1) % mod;
    vector<map<LL, LL> > tct(ct.size() + c[i]);
    forab (j, 0, c[i]) {
      forn (k, ct.size()) for (auto it : ct[k]) {
        LL val = (LL)j + (LL)k - s2;
        LL lim = (val + 100000L * (i + 1)) / (i + 1) - 100000;
        tct[j + k][min(it.first, lim)] = (tct[j + k][min(it.first, lim)] + it.second) % mod;
      }
    }

    ct = tct;
  }
  */

  LL low = 1e6;
  LL tmp1 = 0, tmp2 = 0;
  forn (i, n) {
    tmp1 = (tmp1 + b[i]) % mod;
    tmp2 = (tmp2 + tmp1) % mod;
    low = min(low, (-tmp2 + 100000L * (i + 1)) / (i + 1) - 100000);
  }

  LL high = 1e6;
  LL sm = 0;
  tmp1 = 0, tmp2 = 0;
  forn (i, n) {
    tmp1 = (tmp1 + b[i]) % mod;
    tmp2 = (tmp2 + tmp1) % mod;
    sm = (sm + c[i]) % mod;
    // cout << sm << endl;
    high = min(high, (sm - tmp2 + 100000L * (i + 1)) / (i + 1) - 100000);
  }
  // cout << base << endl;
  // base -= 10;
  // cout << low << " " << high << endl;

  vector<LL> ways(200001);

  forab (t, low - 10, high + 10) {
    // cout << t << endl;
    LL x = t;
    LL s1 = 0, s2 = 0;
    vector<LL> ct;
    ct.pb(1);
    forn (i, n) {
      s1 = (s1 + b[i]) % mod;
      s2 = (s2 + s1) % mod;
      vector<LL> tct(ct.size() + c[i]);
      LL st = 0;
      forn (j, ct.size() + c[i]) {
        if (j < ct.size()) {
          st = (st + ct[j]) % mod;
        }
        if (j - c[i] - 1 >= 0) {
          st = (st + mod - ct[j - c[i] - 1]) % mod;
        }
        if ((LL)j - s2 >= (LL)(i + 1) * x) {
          tct[j] = st;
        }
      }

      ct = tct;
    }

    forn (i, ct.size()) {
      ways[x + 100000] = (ways[x + 100000] + ct[i]) % mod;
    }
    // cout << x << " " << ways[x + 100000] << endl;
  }

  forabd (i, 0, ways.size() - 1) {
    if (ways[i] == 0) {
      ways[i] = ways[i + 1];
    }
  }

  // forab (i, -10, 10) {
  //   cout << i << " " << ways[i + 100000] << endl;
  // }

  int q;
  cin >> q;

  forn (t, q) {
    int x;
    scanf("%d", &x);

    cout << ways[x + 100000] << endl;
  }

  return 0;
}