#include <bits/stdc++.h>

namespace oct {

template <typename _Tp>
_Tp &mini(_Tp &x, _Tp y) {
    return x = std::min(x, y);
}
template <typename _Tp>
_Tp &maxi(_Tp &x, _Tp y) {
    return x = std::max(x, y);
}

#if __cplusplus >= 201103L

template <typename _Tp>
_Tp &mini(_Tp &x, std::initializer_list<_Tp> il) {
    return x = std::min(x, std::min(il));
}

template <typename _Tp>
_Tp &maxi(_Tp &x, std::initializer_list<_Tp> il) {
    return x = std::max(x, std::max(il));
}

#endif

}  // namespace oct

using int64 = long long;

const int P = 1000000007;
const int N = 105;
const int M = N * N;

int n, xl, xr, q, x, ps;
int sb[N], c[N], b[N];

int64 dp[N][M], f[N][M];

std::map<int, int> map;

int solve(int x) {
    memset(dp, 0, sizeof(dp));
    memset(f, 0, sizeof(f));
    dp[0][0] = 1;
    for (int i = 0; i < M; ++i) f[0][i] = 1;
    for (int i = 1; i <= n; ++i) {
        for (int j = std::max(i * x + sb[i], 0); j < M; ++j) {
            dp[i][j] = f[i - 1][j];
            if (j - c[i] > 0) dp[i][j] -= f[i - 1][j - c[i] - 1];
            f[i][j] = j ? f[i][j - 1] + dp[i][j] : dp[i][j];
            dp[i][j] %= P, f[i][j] %= P;
        }
    }
    int64 ans = 0;
    for (int i = 0; i < M; ++i) ans += dp[n][i];
    return (ans % P + P) % P;
}

int main() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0), std::cout.tie(0);
    std::cin >> n;
    ps = 1;
    for (int i = 1; i <= n; ++i) std::cin >> c[i], ps = ps * (c[i] + 1ll) % P;
    for (int i = 1; i < n; ++i) std::cin >> b[i];
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j < i; j++) sb[i] += (i - j) * b[j];
    }
    xl = xr = P;
    for (int i = 1; i <= n; ++i) {
        oct::mini(xl, -((sb[i] - 1) / i + 1));
        oct::mini(xr, (n * N - sb[i] - 1) / i + 1);
    }
    for (int i = xl; i <= xr; ++i) map[i] = solve(i);
    std::cin >> q;
    while (q--) {
        std::cin >> x;
        if (x < xl) std::cout << ps << '\n';
        if (x > xr) std::cout << 0 << '\n';
        if (x >= xl && x <= xr) std::cout << map[x] << '\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>

namespace oct {

template <typename _Tp>
_Tp &mini(_Tp &x, _Tp y) {
    return x = std::min(x, y);
}
template <typename _Tp>
_Tp &maxi(_Tp &x, _Tp y) {
    return x = std::max(x, y);
}

#if __cplusplus >= 201103L

template <typename _Tp>
_Tp &mini(_Tp &x, std::initializer_list<_Tp> il) {
    return x = std::min(x, std::min(il));
}

template <typename _Tp>
_Tp &maxi(_Tp &x, std::initializer_list<_Tp> il) {
    return x = std::max(x, std::max(il));
}

#endif

}  // namespace oct

using int64 = long long;

const int P = 1000000007;
const int N = 105;
const int M = N * N;

int n, xl, xr, q, x, ps;
int sb[N], c[N], b[N];

int64 dp[N][M], f[N][M];

std::map<int, int> map;

int solve(int x) {
    memset(dp, 0, sizeof(dp));
    memset(f, 0, sizeof(f));
    dp[0][0] = 1;
    for (int i = 0; i < M; ++i) f[0][i] = 1;
    for (int i = 1; i <= n; ++i) {
        for (int j = std::max(i * x + sb[i], 0); j < M; ++j) {
            dp[i][j] = f[i - 1][j];
            if (j - c[i] > 0) dp[i][j] -= f[i - 1][j - c[i] - 1];
            f[i][j] = j ? f[i][j - 1] + dp[i][j] : dp[i][j];
            dp[i][j] %= P, f[i][j] %= P;
        }
    }
    int64 ans = 0;
    for (int i = 0; i < M; ++i) ans += dp[n][i];
    return (ans % P + P) % P;
}

int main() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0), std::cout.tie(0);
    std::cin >> n;
    ps = 1;
    for (int i = 1; i <= n; ++i) std::cin >> c[i], ps = ps * (c[i] + 1ll) % P;
    for (int i = 1; i < n; ++i) std::cin >> b[i];
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j < i; j++) sb[i] += (i - j) * b[j];
    }
    xl = xr = P;
    for (int i = 1; i <= n; ++i) {
        oct::mini(xl, -((sb[i] - 1) / i + 1));
        oct::mini(xr, (n * N - sb[i] - 1) / i + 1);
    }
    for (int i = xl; i <= xr; ++i) map[i] = solve(i);
    std::cin >> q;
    while (q--) {
        std::cin >> x;
        if (x < xl) std::cout << ps << '\n';
        if (x > xr) std::cout << 0 << '\n';
        if (x >= xl && x <= xr) std::cout << map[x] << '\n';
    }
    return 0;
}
```