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

#define fi first
#define se second
#define mp make_pair
#define eb emplace_back
typedef long long ll;
typedef pair<int, int> pi;

int n;
const int maxn = 105;

int c[maxn], b[maxn];

ll pb[maxn], ppb[maxn]; 
ll dp[maxn][maxn * maxn];
ll pre[maxn][maxn * maxn];
ll S;
const int mod = 1e9 + 7;

void add(ll &a, ll b){
    a += b;
    if(a >= mod)
        a -= mod;
}

const int offset = 1e4 + 5;
ll L, R;
ll ans[offset * 2];

ll get(ll x){
    memset(dp, 0, sizeof(dp));
    memset(pre, 0, sizeof(pre));
    dp[0][0] = 1;
    for(int i = 0;i < n;i++){
        for(int j = 0;j <= S;j++){
            pre[i][j] = dp[i][j];
            if(j > 0)
                add(pre[i][j], pre[i][j-1]);
        }
        ll lb = x * (i + 1) + ppb[i];
        for(ll j = max(0LL, lb);j <= S;j++){
            ll l = max(x * i + ((i > 0) ? ppb[i-1] : 0), 1LL * j - c[i+1]);  
            ll tmp;
            if(l <= j)
                tmp = (pre[i][j] - ((l > 0) ? pre[i][l-1] : 0) + mod) % mod; 
            else
                tmp = 0;
            add(dp[i+1][j], tmp); 
        }
    }
    ll ans = 0;
    for(int j = 0;j <= S;j++){
        add(ans, dp[n][j]);
    }
    return ans;
}

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    S = 0;
    c[0] = 0;
    for(int i = 1;i <= n;i++){
        cin >> c[i];
        S += c[i];
    }
    b[0] = 0;
    pb[0] = 0;
    ppb[0] = 0;
    for(int i = 1;i < n;i++){
        cin >> b[i];
        pb[i] = pb[i-1] + b[i];
        ppb[i] = ppb[i-1] + pb[i];
    }
    int Q;
    cin >> Q;
    R = (S - ppb[n-1]) / n;
    L = -ppb[n-1] / n;
    for(ll x = L - 5;x <= R + 5;x++){
        ans[x + offset] = get(x); 
    }
    for(int i = 1;i <= Q;i++){
        ll x;
        cin >> x;
        if(x < L - 5){
            cout << ans[L - 5 + offset] << endl;
        }else if(x > R + 5){
            cout << 0 << endl;
        }else{
            cout << ans[x + offset] << endl;
        }
    }
    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;

#define fi first
#define se second
#define mp make_pair
#define eb emplace_back
typedef long long ll;
typedef pair<int, int> pi;

int n;
const int maxn = 105;

int c[maxn], b[maxn];

ll pb[maxn], ppb[maxn]; 
ll dp[maxn][maxn * maxn];
ll pre[maxn][maxn * maxn];
ll S;
const int mod = 1e9 + 7;

void add(ll &a, ll b){
    a += b;
    if(a >= mod)
        a -= mod;
}

const int offset = 1e4 + 5;
ll L, R;
ll ans[offset * 2];

ll get(ll x){
    memset(dp, 0, sizeof(dp));
    memset(pre, 0, sizeof(pre));
    dp[0][0] = 1;
    for(int i = 0;i < n;i++){
        for(int j = 0;j <= S;j++){
            pre[i][j] = dp[i][j];
            if(j > 0)
                add(pre[i][j], pre[i][j-1]);
        }
        ll lb = x * (i + 1) + ppb[i];
        for(ll j = max(0LL, lb);j <= S;j++){
            ll l = max(x * i + ((i > 0) ? ppb[i-1] : 0), 1LL * j - c[i+1]);  
            ll tmp;
            if(l <= j)
                tmp = (pre[i][j] - ((l > 0) ? pre[i][l-1] : 0) + mod) % mod; 
            else
                tmp = 0;
            add(dp[i+1][j], tmp); 
        }
    }
    ll ans = 0;
    for(int j = 0;j <= S;j++){
        add(ans, dp[n][j]);
    }
    return ans;
}

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    S = 0;
    c[0] = 0;
    for(int i = 1;i <= n;i++){
        cin >> c[i];
        S += c[i];
    }
    b[0] = 0;
    pb[0] = 0;
    ppb[0] = 0;
    for(int i = 1;i < n;i++){
        cin >> b[i];
        pb[i] = pb[i-1] + b[i];
        ppb[i] = ppb[i-1] + pb[i];
    }
    int Q;
    cin >> Q;
    R = (S - ppb[n-1]) / n;
    L = -ppb[n-1] / n;
    for(ll x = L - 5;x <= R + 5;x++){
        ans[x + offset] = get(x); 
    }
    for(int i = 1;i <= Q;i++){
        ll x;
        cin >> x;
        if(x < L - 5){
            cout << ans[L - 5 + offset] << endl;
        }else if(x > R + 5){
            cout << 0 << endl;
        }else{
            cout << ans[x + offset] << endl;
        }
    }
    return 0;
}


```