This is the easy version of the problem. The only difference is that in this version q = 1. 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 (q=1).

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
1
-1


Output


56

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.
#include <bits/stdc++.h>
using namespace std;
//using namespace atcoder;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; }
#define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} }
#define pb push_back
#define fi first
#define se second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p){ cout << '(' << p.first << ',' << p.second << ')'; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void chOut(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using v_bool = vector<bool>;
using v_Pii = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;

template<long long MOD>
struct ModInt{

    using ll = long long;
    ll val;

    void setval(ll v) { val = v % MOD; };
    ModInt(): val(0) {}
    ModInt(ll v) { setval(v); };

    ModInt operator+(const ModInt &x) const { return ModInt(val + x.val); }
    ModInt operator-(const ModInt &x) const { return ModInt(val - x.val + MOD); }
    ModInt operator*(const ModInt &x) const { return ModInt(val * x.val); }
    ModInt operator/(const ModInt &x) const { return *this * x.inv(); }
    ModInt operator-() const { return ModInt(MOD - val); }
    ModInt operator+=(const ModInt &x) { return *this = *this + x; }
    ModInt operator-=(const ModInt &x) { return *this = *this - x; }
    ModInt operator*=(const ModInt &x) { return *this = *this * x; }
    ModInt operator/=(const ModInt &x) { return *this = *this / x; }
    bool operator==(const ModInt &x) const { return (*this).val == x.val; }

    friend ostream& operator<<(ostream &os, const ModInt &x) { os << x.val; return os; }
    friend istream& operator>>(istream &is, ModInt &x) { is >> x.val; x.val = (x.val % MOD + MOD) % MOD; return is; }

    ModInt pow(ll n) const {
        ModInt a = 1;
        if(n == 0) return a;
        int i0 = 64 - __builtin_clzll(n);
        for(int i = i0 - 1; i >= 0; i--){
            a = a * a;
            if((n >> i) & 1) a *= (*this); 
        }
        return a;
    }
    ModInt inv() const { return this->pow(MOD - 2); }
};

using mint = ModInt<mod>; mint pow(mint x, long long n) { return x.pow(n); }
//using mint = double; //for debug
using mvec = vector<mint>;
using mmat = vector<mvec>;

struct Combination{

    vector<mint> fact, invfact;

    Combination(int N){
        fact = vector<mint>({mint(1)});
        invfact = vector<mint>({mint(1)});
        fact_initialize(N);
    }

    void fact_initialize(int N){
        int i0 = fact.size();
        if(i0 >= N + 1) return;
        fact.resize(N + 1);
        invfact.resize(N + 1);
        for(int i = i0; i <= N; i++) fact[i] = fact[i - 1] * i;
        invfact[N] = (mint)1 / fact[N];
        for(int i = N - 1; i >= i0; i--) invfact[i] = invfact[i + 1] * (i + 1); 
    }

    mint nCr(int n, int r){
        if(n < 0 || r < 0 || r > n) return mint(0);
        if(fact.size() < n + 1) fact_initialize(n);
        return fact[n] * invfact[r] * invfact[n - r];
    }

    mint nPr(int n, int r){
        if(n < 0 || r < 0 || r > n) return mint(0);
        if(fact.size() < n + 1) fact_initialize(n);
        return fact[n] * invfact[n - r];
    }

    mint Catalan(int n){
        if(n < 0) return 0;
        else if(n == 0) return 1;
        if(fact.size() < 2 * n + 1) fact_initialize(2 * n);
        return fact[2 * n] * invfact[n + 1] * invfact[n];
    }

};

signed main(){

    int n; cin >> n;
    vec c(n); cin >> c;
    vec b(n - 1); cin >> b;
    int q; cin >> q;
    assert(q == 1);
    int x; cin >> x;

    vec p(n);
    p[n - 1] = 0;
    IREP(i, n - 1) p[i] = p[i + 1] + b[i];

    const int M = 600000;
    int th = max(x + p[0], 0LL);
    mvec dp(M + 1, 0);
    FOR(j, p[0], c[0] + p[0] + 1) if(j >= th) dp[j] = 1;
    FOR(i, 1, n){
        FOR(j, 1, M + 1) dp[j] += dp[j - 1];
        IREP(j, M + 1){
            int jmin = j - (p[i] + c[i]), jmax = j - p[i];
            mint tmp = 0;
            if(jmax >= 0) tmp += dp[jmax];
            if(jmin - 1 >= 0) tmp -= dp[jmin - 1];
            dp[j] = tmp;
        }
        REP(j, (i + 1) * th) if(j <= M) dp[j] = 0;
    }
    mint ans = 0;
    REP(j, M + 1) ans += dp[j];
    Out(ans);

    return 0;
}
