#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const ll MOD = 1000000007ll;

int main(){
    ios::sync_with_stdio(0);
    int n;
    cin >> n;
    vector<int> c(n),b(n-1);
    for(auto &it:c)
        cin >> it;
    for(auto &it:b)
        cin >> it;
    vector<ll> bp(n);
    bp[0] = 0ll;
    ll bsum = 0;
    for(int i = 1; i<n; i++){
        bsum += b[i-1];
        bp[i] = (bp[i-1]+bsum);
    }
    int q;
    cin >> q;
    vector<int> qx(q);
    for(auto &it:qx) cin>>it;
    set<int> qset;
    for(auto it:qx) qset.insert(it);
    vector<int> x;
    for(auto it:qset) x.push_back(it);
    vector<ll> dp(n*102,0),dp2;
    map<int,ll> ansmap;
    dp[0] = 1;
    int z = x.size()-1;
    for(int i = 0; i<n; i++){
        for(; x[z]*(i+1)+bp[i]>0; z--){
            ll tgt = x[z]*(i+1ll)+bp[i];
            if(tgt>=n*102){
                ansmap[x[z]] = 0;
                continue;
            }
            dp2 = dp;
            bool failed = false;
            for(int j = i; j<n; j++){
                ll tgt = x[z]*(j+1ll)+bp[j];
                if(tgt>=102*n){
                    ansmap[x[z]] = 0;
                    failed = true;
                    break;
                }
                ll csum = 0;
                for(int k = n*102-1; k>=tgt; k--){
                    if(k-c[j]>=0)
                        csum=(csum+dp2[k-c[j]])%MOD;
                    ll tmp = dp2[k];
                    dp2[k]=(csum)%MOD;
                    csum=(csum-tmp+MOD)%MOD;
                }
                for(int k = 0; k<tgt; k++)
                    dp2[k] = 0;
            }
            ll ans = 0;
            if(!failed){
                for(int j = 0; j<n*102; j++)
                    ans = (ans+dp2[j])%MOD;
                ansmap[x[z]] = ans;
            }
        }
        ll csum = 0;
        for(int k = n*102-1; k>=0; k--){
            if(k-c[i]>=0)
                csum=(csum+dp[k-c[i]])%MOD;
            ll tmp = dp[k];
            dp[k]=csum%MOD;
            csum=(csum-tmp+MOD)%MOD;
        }
        csum = 0;
    }
    ll ans = 0;
    for(int j = 0; j<n*102; j++)
        ans = (ans+dp[j])%MOD;
    for(;z>=0;z--)
        ansmap[x[z]] = ans;
    for(auto it:qx)
        cout << ansmap[it] << endl;
    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;
typedef long long ll;
const ll MOD = 1000000007ll;

int main(){
    ios::sync_with_stdio(0);
    int n;
    cin >> n;
    vector<int> c(n),b(n-1);
    for(auto &it:c)
        cin >> it;
    for(auto &it:b)
        cin >> it;
    vector<ll> bp(n);
    bp[0] = 0ll;
    ll bsum = 0;
    for(int i = 1; i<n; i++){
        bsum += b[i-1];
        bp[i] = (bp[i-1]+bsum);
    }
    int q;
    cin >> q;
    vector<int> qx(q);
    for(auto &it:qx) cin>>it;
    set<int> qset;
    for(auto it:qx) qset.insert(it);
    vector<int> x;
    for(auto it:qset) x.push_back(it);
    vector<ll> dp(n*102,0),dp2;
    map<int,ll> ansmap;
    dp[0] = 1;
    int z = x.size()-1;
    for(int i = 0; i<n; i++){
        for(; x[z]*(i+1)+bp[i]>0; z--){
            ll tgt = x[z]*(i+1ll)+bp[i];
            if(tgt>=n*102){
                ansmap[x[z]] = 0;
                continue;
            }
            dp2 = dp;
            bool failed = false;
            for(int j = i; j<n; j++){
                ll tgt = x[z]*(j+1ll)+bp[j];
                if(tgt>=102*n){
                    ansmap[x[z]] = 0;
                    failed = true;
                    break;
                }
                ll csum = 0;
                for(int k = n*102-1; k>=tgt; k--){
                    if(k-c[j]>=0)
                        csum=(csum+dp2[k-c[j]])%MOD;
                    ll tmp = dp2[k];
                    dp2[k]=(csum)%MOD;
                    csum=(csum-tmp+MOD)%MOD;
                }
                for(int k = 0; k<tgt; k++)
                    dp2[k] = 0;
            }
            ll ans = 0;
            if(!failed){
                for(int j = 0; j<n*102; j++)
                    ans = (ans+dp2[j])%MOD;
                ansmap[x[z]] = ans;
            }
        }
        ll csum = 0;
        for(int k = n*102-1; k>=0; k--){
            if(k-c[i]>=0)
                csum=(csum+dp[k-c[i]])%MOD;
            ll tmp = dp[k];
            dp[k]=csum%MOD;
            csum=(csum-tmp+MOD)%MOD;
        }
        csum = 0;
    }
    ll ans = 0;
    for(int j = 0; j<n*102; j++)
        ans = (ans+dp[j])%MOD;
    for(;z>=0;z--)
        ansmap[x[z]] = ans;
    for(auto it:qx)
        cout << ansmap[it] << endl;
    return 0;
}
```