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 ll long long
#define int ll
#define ld long double
#define pii pair<int, int>
#define f first
#define s second
#define boost() cin.tie(0), cin.sync_with_stdio(0)

const int MN = 105, MOD = 1e9 + 7;

int n, a[MN], b[MN], c[MN], psa[MN], psum[MN], q, x, dp[12000005], ndp[12000005], lo_ans = 1;
map<int, int> memo;

int32_t main() {
    boost();

    cin >> n;
    for (int i = 1; i <= n; i++) cin >> c[i], psa[i] = psa[i - 1] + c[i], lo_ans = lo_ans * (c[i] + 1) % MOD;
    for (int i = 1; i < n; i++) cin >> b[i];
    int sum = 0;
    for (int i = 2; i <= n; i++) a[i] = a[i - 1] + b[i - 1];
    int v = 0;
    for (int i = 1; i <= n; i++) sum += a[i], psum[i] = psum[i - 1] + a[i];
    for (int i = 1; i <= n; i++) v = min(v, -(psum[i] / i));
    //printf("%lld\n", sum);
    //if (x > c[1]) return 0 * printf("0\n");
    cin >> q;
    while (q--) {
        cin >> x;
        if (x < v - 100) printf("%lld\n", lo_ans);
        else if (x > v + 100) printf("0\n");
        else if (memo.count(x)) printf("%lld\n", memo[x]);
        else {
            memset(dp, 0, sizeof(dp));
            memset(ndp, 0, sizeof(ndp));
            int ans = 0;
            dp[0] = 1;
            for (int i = 1; i <= n; i++) {
                for (int j = 0; j <= psa[i - 1]; j++) {
                    int lo = 0, hi = c[i];
                    //if (i == 1) lo = max(lo, x);
                    ndp[j + lo] += dp[j];
                    ndp[j + lo] %= MOD;
                    ndp[j + hi + 1] -= dp[j];
                    ndp[j + hi + 1] %= MOD;
                }
                for (int j = 1; j <= psa[i]; j++) {
                    ndp[j] += ndp[j - 1];
                    ndp[j] %= MOD;
                }
                for (int j = 0; j <= psa[i]; j++) {
                    dp[j] = ndp[j];
                    if (j < psum[i] + x * i) dp[j] = 0;
                }
                //printf("\n");
                for (int j = 0; j <= psa[i] + 1; j++) {
                    ndp[j] = 0;
                }
            }
            for (int i = 0; i <= psa[n]; i++) ans += dp[i], ans %= MOD;
            ans %= MOD, ans += MOD, ans %= MOD;
            printf("%lld\n", memo[x] = ans);
        }
    }

    return 0;
}