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

const int MOD = 1e9 + 7;
int c[105], d[105], sd[105], dp[105][10005];

int solve(int n, int x) {
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    int res = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= 100 * n; j++) {
            if (j) dp[i][j] = dp[i][j - 1];
            dp[i][j] += dp[i - 1][j];
            dp[i][j] %= MOD;
            if (j > c[i]) dp[i][j] -= dp[i - 1][j - c[i] - 1];

            dp[i][j] %= MOD;
            
        }
        for (int j = 0; j <= 100 * n; j++) {
            if (j < i * x - sd[i]) dp[i][j] = 0;
            else if (i == n) {
                res += dp[i][j];
                res %= MOD;
            }
        }
    }

    return (res + MOD) % MOD;
}

int main() {
    // freopen("input.txt", "r", stdin);
    ios::sync_with_stdio(0);
    cin.tie(NULL);

    int n;
    cin >> n;

    for (int i = 1; i <= n; i++) {
        cin >> c[i];
    }    

    for (int i = 2; i <= n; i++) {
        int b;
        cin >> b;
        d[i] = d[i - 1] - b;
        sd[i] = sd[i - 1] + d[i];
    }

    // now, we have d[i] <= a[i] <= c[i] + d[i]
    // we need a[1] <= a[2] <= .... <= a[n]
    // a[1] = min(s[i] / i), s is prefix sum
    // s[i] >= i * x, for all i

    // now, let a[i] = a[i] + d[i]
    // then, s[i] >= i * x - sd[i]
    int q, x;
    cin >> q;

    // cout << solve(n, x) << endl;  
    int minSD = *min_element(sd, sd + n + 1);
    int minX = minSD / n - 1;
    int maxX = minSD / n + 100 + 1;

    vector<int> preCalc(105);
    for (int i = minX; i <= maxX; i++) {
        preCalc[i - minX] = solve(n, i);
    }

    for (int i = 1; i <= q; i++) {
        int x;
        cin >> x;
        if (x < minX) {
            cout << preCalc[0] << "\n";
        }
        else if (x > maxX) {
            cout << 0 << "\n";
        }
        else {
            cout << preCalc[x - minX] << "\n";
        }
    }

    return 0;
;}
### Prompt

Your task is to create 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;
int c[105], d[105], sd[105], dp[105][10005];

int solve(int n, int x) {
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    int res = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= 100 * n; j++) {
            if (j) dp[i][j] = dp[i][j - 1];
            dp[i][j] += dp[i - 1][j];
            dp[i][j] %= MOD;
            if (j > c[i]) dp[i][j] -= dp[i - 1][j - c[i] - 1];

            dp[i][j] %= MOD;
            
        }
        for (int j = 0; j <= 100 * n; j++) {
            if (j < i * x - sd[i]) dp[i][j] = 0;
            else if (i == n) {
                res += dp[i][j];
                res %= MOD;
            }
        }
    }

    return (res + MOD) % MOD;
}

int main() {
    // freopen("input.txt", "r", stdin);
    ios::sync_with_stdio(0);
    cin.tie(NULL);

    int n;
    cin >> n;

    for (int i = 1; i <= n; i++) {
        cin >> c[i];
    }    

    for (int i = 2; i <= n; i++) {
        int b;
        cin >> b;
        d[i] = d[i - 1] - b;
        sd[i] = sd[i - 1] + d[i];
    }

    // now, we have d[i] <= a[i] <= c[i] + d[i]
    // we need a[1] <= a[2] <= .... <= a[n]
    // a[1] = min(s[i] / i), s is prefix sum
    // s[i] >= i * x, for all i

    // now, let a[i] = a[i] + d[i]
    // then, s[i] >= i * x - sd[i]
    int q, x;
    cin >> q;

    // cout << solve(n, x) << endl;  
    int minSD = *min_element(sd, sd + n + 1);
    int minX = minSD / n - 1;
    int maxX = minSD / n + 100 + 1;

    vector<int> preCalc(105);
    for (int i = minX; i <= maxX; i++) {
        preCalc[i - minX] = solve(n, i);
    }

    for (int i = 1; i <= q; i++) {
        int x;
        cin >> x;
        if (x < minX) {
            cout << preCalc[0] << "\n";
        }
        else if (x > maxX) {
            cout << 0 << "\n";
        }
        else {
            cout << preCalc[x - minX] << "\n";
        }
    }

    return 0;
;}
```