#include <bits/stdc++.h>

typedef long long ll;
using namespace std;
const int MOD = 1e9+7;
ll dp[101][10011], ans[202000];
int a[101], b[101], n;

void solve(int x){
    int X = x;
    ll S = x;
    for (int i=0;i<=n;i++) fill(dp[i], dp[i]+10011, 0);
    for (int j=max(x, 0);j<=a[0];j++) dp[0][j] = 1;
    for (int i=1;i<n;i++){
        x += b[i-1];
        S += x;
        ll cur = 0;
        for (int k=0;k<=a[i];k++){
            if (max(S, 0LL)-k<0) break;
            cur += dp[i-1][max(S, 0LL)-k];
            if (cur>=MOD) cur -= MOD;
        }
        for (int j=max(S, 0LL);j<10010;j++){
            dp[i][j] = cur;
            cur += dp[i-1][j+1];
            if (j-a[i]>=0) cur -= dp[i-1][j-a[i]];
            cur %= MOD;
            if (cur<0) cur += MOD;
        }
    }
    for (int j=0;j<10010;j++){
        ans[X+100100] += dp[n-1][j];
        if (ans[X+100100]>=MOD) ans[X+100100] -= MOD;
    }
}

int main(){
    scanf("%d", &n);
    ll tmp2 = 0;
    for (int i=0;i<n;i++){
        scanf("%d", a+i);
        tmp2 += a[i];
    }
    for (int i=0;i<n-1;i++){
        scanf("%d", b+i);
    }
    int x = 100100;
    ll S = x, X = x;
    for (int i=1;i<n;i++){
        X += b[i-1];
        S += X;
    }
    while(S>tmp2){
        x--; S -= n;
    }
    for (int i=0;i<111;i++) solve(x-i);
    x -= 111;
    for (;x+100100>=0;x--) ans[x+100100] = ans[x+100101];

    int q;
    scanf("%d", &q);
    while(q--){
        int tmp;
        scanf("%d", &tmp);
        printf("%lld\n", ans[tmp+100100]);
    }
    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>

typedef long long ll;
using namespace std;
const int MOD = 1e9+7;
ll dp[101][10011], ans[202000];
int a[101], b[101], n;

void solve(int x){
    int X = x;
    ll S = x;
    for (int i=0;i<=n;i++) fill(dp[i], dp[i]+10011, 0);
    for (int j=max(x, 0);j<=a[0];j++) dp[0][j] = 1;
    for (int i=1;i<n;i++){
        x += b[i-1];
        S += x;
        ll cur = 0;
        for (int k=0;k<=a[i];k++){
            if (max(S, 0LL)-k<0) break;
            cur += dp[i-1][max(S, 0LL)-k];
            if (cur>=MOD) cur -= MOD;
        }
        for (int j=max(S, 0LL);j<10010;j++){
            dp[i][j] = cur;
            cur += dp[i-1][j+1];
            if (j-a[i]>=0) cur -= dp[i-1][j-a[i]];
            cur %= MOD;
            if (cur<0) cur += MOD;
        }
    }
    for (int j=0;j<10010;j++){
        ans[X+100100] += dp[n-1][j];
        if (ans[X+100100]>=MOD) ans[X+100100] -= MOD;
    }
}

int main(){
    scanf("%d", &n);
    ll tmp2 = 0;
    for (int i=0;i<n;i++){
        scanf("%d", a+i);
        tmp2 += a[i];
    }
    for (int i=0;i<n-1;i++){
        scanf("%d", b+i);
    }
    int x = 100100;
    ll S = x, X = x;
    for (int i=1;i<n;i++){
        X += b[i-1];
        S += X;
    }
    while(S>tmp2){
        x--; S -= n;
    }
    for (int i=0;i<111;i++) solve(x-i);
    x -= 111;
    for (;x+100100>=0;x--) ans[x+100100] = ans[x+100101];

    int q;
    scanf("%d", &q);
    while(q--){
        int tmp;
        scanf("%d", &tmp);
        printf("%lld\n", ans[tmp+100100]);
    }
    return 0;
}

```