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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

#define u_map unordered_map
#define u_set unordered_set
#define u_multiset unordered_multiset

using ll = long long;
using vvi = vector<vector<int>>;
using vi = vector<int>;
using vvll = vector<vector<long long>>;
using vll = vector<long long>;
using vd = vector<double>;
using vvd = vector<vector<double>>;
using pii = pair<int, int>;
using vpii = vector<pair<int, int>>;

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef DEBUG
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

const ll MOD = 1e9 + 7;
const int T = 1e4 + 5;
// const int T = 20;

ll mul(ll a, ll b){
    return a * b % MOD;
}

ll add(ll a, ll b){
    return (a + b) % MOD;
}

ll neg(ll x){
    return (MOD - x) % MOD;
}

ll calc(int n, vi &p_pref, int x, vi &c){

    vi threshold(n);
    for(int i = 0; i < n; i++){
        threshold[i] = (i+1) * x + p_pref[i+1];
    }
    debug(threshold);

    vvll dp(n+1, vll(T));
    vvll dpp(n+1, vll(T+1));
    dp[0][0] = 1;
    for(int i = 1; i <= T; i++){
        dpp[0][i] = 1;
    }
    for(int i = 1; i <= n; i++){
        for(int j = max(0, threshold[i-1]); j < T; j++){
            int fi = max(0, j-c[i-1]);
            // for(int k = 0; k <= min(j, c[i-1]); k++){
            //     dp[i][j] = add(dp[i][j], dp[i-1][j-k]);
            // }
            dp[i][j] = add(dpp[i-1][j+1], neg(dpp[i-1][fi]));
        }
        for(int j = 1; j <= T; j++){
            dpp[i][j] = add(dpp[i][j-1], dp[i][j-1]);
        }
    }
    ll ans = 0;
    for(int j = 0; j < T; j++){
        ans = add(ans, dp[n][j]);
    }
    return ans;
}

void solve(){
    int n;
    cin >> n;
    vi c(n), b(n-1), p(n);
    for(int i = 0; i < n; i++) cin >> c[i];
    for(int i = 0; i < n-1; i++) cin >> b[i];
    for(int i = 1; i < n; i++){
        p[i] = p[i-1] + b[i-1];
    }
    vi p_pref(n+1);
    for(int i = 1; i < n+1; i++){
        p_pref[i] = p[i-1] + p_pref[i-1];
    }

    // precalc
    int start = (102 * n - p_pref[n]) / n;
    int end;
    map<int, ll> ans;
    for(int i = 0; i < 105; i++){
        int val = start - i;
        ll qa = calc(n, p_pref, val, c);
        ans[val] = qa;
        end = val;
    }

    ll total = 1;
    for(int i = 0; i < n; i++){
        total = mul(total, c[i] + 1);
    }

    int q;
    cin >> q;
    while(q--){
        int x;
        cin >> x;
        if(x > start){
            cout << 0 << '\n';
        }
        else if(x < end){
            cout << total << '\n';
        }
        else{
            cout << ans[x] << '\n';
        }
    }
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}