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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
#include <fstream>
#include <array>
using namespace std;
#define int long long
#define ii pair <int, int>
#define app push_back
#define all(a) a.begin(), a.end()
#define bp __builtin_popcountll
#define ll long long
#define mp make_pair
#define x first
#define y second
#define Time (double)clock()/CLOCKS_PER_SEC
#define debug(x) std::cerr << #x << ": " << x << '\n';

#define FORI(i,a,b) for (int i = (a); i < (b); ++i)
#define FOR(i,a) FORI(i,0,a)
#define ROFI(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define ROF(i,a) ROFI(i,0,a)
#define rep(a) FOR(_,a)
#define each(a,x) for (auto& a: x)
#define FORN(i, n) FORI(i, 1, n + 1)

using vi = vector<int>;
template <typename T>
std::istream& operator >>(std::istream& input, std::pair <T, T> & data)
{
    input >> data.x >> data.y;
    return input;
}
template <typename T>
std::istream& operator >>(std::istream& input, std::vector<T>& data)
{
    for (T& x : data)
        input >> x;
    return input;
}
template <typename T>
std::ostream& operator <<(std::ostream& output, const pair <T, T> & data)
{
    output << "(" << data.x << "," << data.y << ")";
    return output;
}
template <typename T>
std::ostream& operator <<(std::ostream& output, const std::vector<T>& data)
{
    for (const T& x : data)
        output << x << " ";
    return output;
}
std::ostream& operator <<(std::ostream& output, const __int128 &x)
{
    __int128 n = x;
    if (n == 0) {
        output << "0";
        return output;
    }
    if (n < 0) {
        n = -n;
        output << "-";
    }
    string s;
    while (n) {
        s += '0' + (n%10);
        n /= 10;
    }
    reverse(all(s));
    cout << s;
    return output;
}
ll div_up(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll div_down(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down 
ll math_mod(ll a, ll b) { return a - b * div_down(a, b); }
#define tcT template<class T
#define tcTU tcT, class U
tcT> using V = vector<T>; 
tcT> bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0; 
} // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0; 
}
ll gcd(ll a, ll b) {
    while (b) {
        tie(a, b) = mp(b, a % b);
    }
    return a;
}
int Bit(int mask, int bit) { return (mask >> bit) & 1; }

template<int MOD, int RT> struct mint {
    static const int mod = MOD;
    static constexpr mint rt() { return RT; } // primitive root for FFT
    int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
    mint() { v = 0; }
    mint(ll _v) { v = (int)((-MOD < _v && _v < MOD) ? _v : _v % MOD);
        if (v < 0) v += MOD; }
    friend bool operator==(const mint& a, const mint& b) { 
        return a.v == b.v; }
    friend bool operator!=(const mint& a, const mint& b) { 
        return !(a == b); }
    friend bool operator<(const mint& a, const mint& b) { 
        return a.v < b.v; }
    friend string ts(mint a) { return to_string(a.v); }

    mint& operator+=(const mint& m) { 
        if ((v += m.v) >= MOD) v -= MOD; 
        return *this; }
    mint& operator-=(const mint& m) { 
        if ((v -= m.v) < 0) v += MOD; 
        return *this; }
    mint& operator*=(const mint& m) { 
        v = (int)((ll)v*m.v%MOD); return *this; }
    mint& operator/=(const mint& m) { return (*this) *= inv(m); }
    friend mint pow(mint a, ll p) {
        mint ans = 1; assert(p >= 0);
        for (; p; p /= 2, a *= a) if (p&1) ans *= a;
        return ans; }
    mint & operator ^=(const int &p) { return (*this) = pow(this, p); }

    friend mint inv(const mint& a) { assert(a.v != 0); 
        return pow(a,MOD-2); }
        
    mint operator-() const { return mint(-v); }
    mint& operator++() { return *this += 1; }
    mint& operator--() { return *this -= 1; }
    friend mint operator+(mint a, const mint& b) { return a += b; }
    friend mint operator-(mint a, const mint& b) { return a -= b; }
    friend mint operator*(mint a, const mint& b) { return a *= b; }
    friend mint operator/(mint a, const mint& b) { return a /= b; }
    friend mint operator^(mint a, const int p) { return pow(a, p); }
};

const int MOD = 1e9+7;

typedef mint<MOD,5> mi; // 5 is primitive root for both common mods
typedef vector<mi> vmi;

std::ostream& operator << (std::ostream& o, const mi& a)
{
    cout << a.v;
    return o;
}

vector<vmi> scmb; // small combinations
void genComb(int SZ) {
    scmb.assign(SZ,vmi(SZ)); scmb[0][0] = 1;
    FORI(i,1,SZ) FOR(j,i+1) 
        scmb[i][j] = scmb[i-1][j]+(j?scmb[i-1][j-1]:0);
}
 
vmi invs, fac, ifac; // make sure to convert to LL before doing any multiplications ...
void genFac(int SZ) {
    invs.resize(SZ), fac.resize(SZ), ifac.resize(SZ); 
    invs[1] = fac[0] = ifac[0] = 1; 
    FORI(i,2,SZ) invs[i] = mi(-(ll)MOD/i*(int)invs[MOD%i]);
    FORI(i,1,SZ) {
        fac[i] = fac[i-1]*i;
        ifac[i] = ifac[i-1]*invs[i];
    }
}
mi comb(int a, int b) {
    if (a < b || b < 0) return 0;
    assert(a < fac.size());
    return fac[a]*ifac[b]*ifac[a-b];
}

const int N = 107;
int n, c[N], b[N], pb[N], ppb[N];

mi dp[N][N * N];

const int C = 2e5+7, S = 100 * 1000;

bool used[C];
mi res[C];

signed main() {
    #ifdef LOCAL
    #else
    #define endl '\n'
    ios_base::sync_with_stdio(0); cin.tie(0);
    #endif
    cin >> n;
    FOR (i, n) {
        cin >> c[i];
    }
    FOR (i, n - 1) {
        cin >> b[i];
        pb[i + 1] = pb[i] + b[i];
    }

    FORN (i, n - 1) {
        ppb[i] = ppb[i - 1] + pb[i];
    }

    mi al = 1;
    FOR (i, n) {
        al *= c[i] + 1;
    }

    int q;
    cin >> q;
    rep (q) {
        int key;
        cin >> key;

        if (key * n < -ppb[n - 1]) {
            cout << al << endl;
            continue;
        }

        if (key * n > 100 * n - ppb[n - 1]) {
            cout << 0 << endl;
            continue;
        }

        if (used[key + S]) {
            cout << res[key + S] << endl;
            continue;
        }

        FOR (i, N) {
            FOR (j, N * N) {
                dp[i][j] = 0;
            }
        }
        dp[0][0] = 1;
        FOR (i, n) {
            FOR (sum, N * N) {
                mi x = dp[i][sum];
                if (x.v) {
                    FOR (add, c[i] + 1) {
                        if (sum+add-ppb[i] >= key*(i + 1)) {
                            dp[i + 1][sum + add] += dp[i][sum];
                        }
                    }
                }
            }
        }
        mi ans = 0;
        FOR (i, N * N) {
            ans += dp[n][i];
        }
        used[key + S] = 1;
        res[key + S] = ans;
        cout << ans << endl;
    }
}