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<algorithm>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#define miu 105
#define ll long long
using namespace std;
const long long mod = 1000000007;
inline int read() {
    register int x = 0, f = 1; register char ch = getchar();
    while(ch < '0' || ch > '9'){if(ch == '-')f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = x * 10 + ch - 48; ch = getchar();}
    return x * f;
}
int n, q;
int b[miu], c[miu];
int sb[miu];
ll dp[110][2][miu * miu], s[miu * miu];
signed main(){
    n = read();
    for(int i = 1; i <= n; ++i) c[i] = read();
    for(int i = 1; i < n; ++i) b[i] = read();
    for(int i = 1; i < n; ++i) sb[i] = sb[i - 1] + b[i];
    for(int i = 1; i < n; ++i) sb[i] += sb[i-1];
    ll pred = 1;
    for(int i = 1; i <= n; ++i) pred = pred * (c[i] + 1) % mod;
    int mn = -sb[n - 1] / n - 1, mx = miu - sb[n - 1] / n + 1;
    for(int i = mn; i <= mx; ++i){
        for(int j = 0; j < miu * miu; ++j) dp[i - mn][(n + 1) % 2][j] = 1;
        for(int j = n; j > 0; --j){
            s[0] = dp[i - mn][(j + 1) % 2][0];
            for(int k = 0; k < miu * miu; ++k)
                s[k] = (s[k - 1] + dp[i - mn][(j + 1) % 2][k]) % mod;
            for(int k = 0; k < miu * miu; ++k){
                int l=max(k,i*j + sb[j - 1]), r = min(k + c[j], miu * miu - 1);
                if(l>r) dp[i - mn][j % 2][k] = 0;
                else dp[i - mn][j % 2][k] = (s[r]-(l ? s[l - 1] : 0)+mod) % mod;
            }
        }
    }
    int q = read();
    while(q--){
        int x = read();
        if(x < mn) printf("%lld\n",pred);
        else if(x > mx) puts("0");
        else printf("%lld\n",dp[x - mn][1][0]);
    }
    return 0;
}