#include <bits/stdc++.h>
using namespace std;

const int MOD = 1e9+7, N = 100, A = 100;
int n, q, b[N], c[N], sol[A+1], dp[2][N*A+1];

void add(int & x, int y) {
    x += y;
    if(x >= MOD) x -= MOD;
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; i++) cin >> c[i];
    for (int i = 0; i < n-1; i++) cin >> b[i];
    int minx = 0, sum = 0, act = 0;
    for (int k = 2; k <= n; k++) {
        act += b[k-2];
        sum -= act;
        minx = min(minx, sum/k);
    }
    for (int x = minx; x <= minx+A; x++) {
        dp[1][0] = 1;
        for (int s = 1; s <= N*A; s++) dp[1][s] = 0;
        int lb = x, act = x;
        for (int i = 0; i < n; i++) {
            dp[i&1][0] = dp[(i&1)^1][0];
            for (int s = 1; s <= N*A; s++) {
                dp[i&1][s] = dp[i&1][s-1];
                add(dp[i&1][s], dp[(i&1)^1][s]);
                if(s > c[i]) add(dp[i&1][s], MOD-dp[(i&1)^1][s-c[i]-1]);
            }
            for (int s = 0; s < min(lb, N*A+1); s++) dp[i&1][s] = 0;
            if(i+1 < n) {
                act += b[i];
                lb += act;
            }
        }
        for (int s = 0; s <= N*A; s++) add(sol[x-minx], dp[(n-1)&1][s]);
    }
    long long all = 1;
    for (int i = 0; i < n; i++) all = all * (c[i]+1) % MOD;

    cin >> q;
    while(q--) {
        int x; cin >> x;
        cout << (x < minx ? all : (x > minx + A ? 0 : sol[x-minx])) << "\n";
    }
    return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

const int MOD = 1e9+7, N = 100, A = 100;
int n, q, b[N], c[N], sol[A+1], dp[2][N*A+1];

void add(int & x, int y) {
    x += y;
    if(x >= MOD) x -= MOD;
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; i++) cin >> c[i];
    for (int i = 0; i < n-1; i++) cin >> b[i];
    int minx = 0, sum = 0, act = 0;
    for (int k = 2; k <= n; k++) {
        act += b[k-2];
        sum -= act;
        minx = min(minx, sum/k);
    }
    for (int x = minx; x <= minx+A; x++) {
        dp[1][0] = 1;
        for (int s = 1; s <= N*A; s++) dp[1][s] = 0;
        int lb = x, act = x;
        for (int i = 0; i < n; i++) {
            dp[i&1][0] = dp[(i&1)^1][0];
            for (int s = 1; s <= N*A; s++) {
                dp[i&1][s] = dp[i&1][s-1];
                add(dp[i&1][s], dp[(i&1)^1][s]);
                if(s > c[i]) add(dp[i&1][s], MOD-dp[(i&1)^1][s-c[i]-1]);
            }
            for (int s = 0; s < min(lb, N*A+1); s++) dp[i&1][s] = 0;
            if(i+1 < n) {
                act += b[i];
                lb += act;
            }
        }
        for (int s = 0; s <= N*A; s++) add(sol[x-minx], dp[(n-1)&1][s]);
    }
    long long all = 1;
    for (int i = 0; i < n; i++) all = all * (c[i]+1) % MOD;

    cin >> q;
    while(q--) {
        int x; cin >> x;
        cout << (x < minx ? all : (x > minx + A ? 0 : sol[x-minx])) << "\n";
    }
    return 0;
}

```