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;
const int c=105, c2=c*c;
int n, t[c], dif[c], mini[c];
long long dp[c][c2], kom[c][c2], sum, q, mod=1000000007;
map<int, long long> valasz;
int main()
{
    ios_base::sync_with_stdio(false);
    cin >> n;
    dp[0][0]=1;
    for (int i=0; i<c2; i++) {
        kom[0][i]=1;
    }
    for (int i=1; i<=n; i++) {
        cin >> t[i];
    }
    for (int i=2; i<=n; i++) {
        cin >> dif[i];
        sum+=dif[i];
        mini[i]=mini[i-1]+sum;
    }
    int kis=-mini[n]/n-10, nagy=kis+220;
    //cout << "hatar " << kis << " " << nagy << "\n";
    for (int ert=kis; ert<=nagy; ert++) {
        for (int i=1; i<=n; i++) {
            int x=t[i];
            int kell=mini[i]+i*ert;
            for (int j=0; j<c2; j++) {
                dp[i][j]=kom[i-1][j];
                if (j>x) {
                    dp[i][j]-=kom[i-1][j-x-1];
                }
                dp[i][j]+=mod, dp[i][j]%=mod;
                if (j<kell) {
                    dp[i][j]=0;
                }
                kom[i][j]=dp[i][j];
                if (j) {
                    kom[i][j]+=kom[i][j-1];
                    kom[i][j]%=mod;
                }
            }
        }
        valasz[ert]=kom[n][c2-1];
    }
    int q;
    cin >> q;
    while (q--) {
        int ert;
        cin >> ert;
        ert=max(kis, min(ert, nagy));
        cout << valasz[ert] << "\n";
    }
    return 0;
}
/*
3
100 100 100
100 100
*/



















/*



#include <bits/stdc++.h>

using namespace std;
const int c=105, c2=c*c;
int n, t[c], dif[c], mini[c];
long long dp[c][c2], kom[c][c2], sum, q, mod=1000000007;
int main()
{
    ios_base::sync_with_stdio(false);
    cin >> n;
    dp[0][0]=1;
    for (int i=0; i<c2; i++) {
        kom[0][i]=1;
    }
    for (int i=1; i<=n; i++) {
        cin >> t[i];
    }
    for (int i=2; i<=n; i++) {
        cin >> dif[i];
        sum+=dif[i];
        mini[i]=mini[i-1]+sum;
    }
    cin >> q;
    while (q--) {
        int ert;
        cin >> ert;

        for (int i=1; i<=n; i++) {
            int x=t[i];
            int kell=mini[i]+i*ert;
            for (int j=0; j<c2; j++) {
                dp[i][j]=kom[i-1][j];
                if (j>x) {
                    dp[i][j]-=kom[i-1][j-x-1];
                }
                dp[i][j]+=mod, dp[i][j]%=mod;
                if (j<kell) {
                    dp[i][j]=0;
                }
                kom[i][j]=dp[i][j];
                if (j) {
                    kom[i][j]+=kom[i][j-1];
                    kom[i][j]%=mod;
                }
            }
        }
        cout << kom[n][c2-1] << "\n";
    }
    return 0;
}
*/