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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int mod = 1000000007;

// returns the floor of the fraction a/b
// assuming b>0
int floor(int a, int b) {
    if (a>=0)
        return a/b;
    return -((-a+b-1)/b);
}

vector<int> psums(vector<int> & bs, int x,int n) {
    vector<int> ps = vector<int>(n,0);
    for (int i=0; i<n; i++) {
        ps[i] = (i+1)*x;
        for (int j = i; j>0; j--) {
            ps[i] = ps[i] + j*bs[i-j];
        }
    }
    return ps;
}

int computeAns(vector<int> & bs, vector<int> & cs, int x, int n, vector<int> & zpsums) {
    // dp[i][j] is the number of a1, dots, ai
    // satisfying the bounds such that a1+dots+ai = j
    // and sp[i][j] is the number of a1, dots, ai
    // satisfying the bounds such that a1+dots+ai <= j
    vector<vector<int>> dp = vector<vector<int>>(n+1,vector<int>(10001,0));
    vector<vector<int>> sp = vector<vector<int>>(n+1,vector<int>(10001,0));
    dp[0][0] = 1;
    sp[0][0] = 1;
    int prevHigh = 0;
    for (int i=1; i<=n; i++) {
        int low = max(0,zpsums[i-1]+i*x);
        int nextHigh = prevHigh + cs[i-1];
        for (int j=low; j<=nextHigh; j++) {
            int smaller = j-cs[i-1]-1>=0 ? sp[i-1][j-cs[i-1]-1] : 0;
            int larger = j>=prevHigh ? sp[i-1][prevHigh] : sp[i-1][j];
            dp[i][j] = (larger-smaller+mod)%mod;
            sp[i][j] = j==low ? dp[i][j] : (sp[i][j-1]+dp[i][j])%mod;
        }
        prevHigh = nextHigh;
    }
    return sp[n][prevHigh];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n; cin >> n;
    vector<int> cs;
    for (int i=0; i<n; i++) {
        int t; cin >>t;
        cs.push_back(t);
    }
    vector<int> bs;
    for (int i=0; i<n-1; i++) {
        int t; cin >> t;
        bs.push_back(t);
    }
    int q; cin >> q;
    vector<int> xs;
    for (int i=0; i<q; i++) {
        int t; cin >> t;
        xs.push_back(t);
    }

    vector<int> zpsums = psums(bs,0,n);
    int minCutoff = 0;
    for (int i=1; i<=n; i++) {
        int next = floor(-zpsums[i-1],i);
        minCutoff = min(minCutoff,next);
    }

    int maxCutoff = cs[0];
    int currSum = cs[0];
    for (int i=2; i<=n; i++) {
        currSum = currSum+cs[i-1];
        int next = floor(currSum-zpsums[i-1],i);
        maxCutoff = min(maxCutoff,next);
    }

    vector<int> ansMemoi = vector<int>(maxCutoff-minCutoff+1,-1);
    
    ansMemoi[0] = computeAns(bs,cs,minCutoff,n,zpsums);

    for (int i=0; i<q; i++) {
        if (xs[i]<= minCutoff) {
            cout << ansMemoi[0] <<'\n';
            continue;
        }
        if (xs[i]>maxCutoff) {
            cout << 0 << '\n';
            continue;
        }
        if (ansMemoi[xs[i]-minCutoff]!= -1) {
            cout << ansMemoi[xs[i]-minCutoff] << '\n';
            continue;
        }
        int ans = computeAns(bs,cs,xs[i],n,zpsums);
        ansMemoi[xs[i]-minCutoff] = ans;
        cout << ans << '\n';
    }
}