Problem: 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.

Solution (in CPP):

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

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
const int N = 105;

const int MOD = (int)1e9 + 7;

struct mi {
    typedef decay<decltype(MOD)>::type T;
    /// don't silently convert to T
    T v; explicit operator T() const { return v; }
    mi() { v = 0; }
    mi(ll _v) {
        v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
        if (v < 0) v += MOD;
    }
    friend bool operator==(const mi& a, const mi& b) {
        return a.v == b.v; }
    friend bool operator!=(const mi& a, const mi& b) {
        return !(a == b); }
    friend bool operator<(const mi& a, const mi& b) {
        return a.v < b.v; }
//    friend void re(mi& a) { ll x; re(x); a = mi(x); }
//    friend str ts(mi a) { return ts(a.v); }

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

    mi operator-() const { return mi(-v); }
    mi& operator++() { return *this += 1; }
    mi& operator--() { return *this -= 1; }
    friend mi operator+(mi a, const mi& b) { return a += b; }
    friend mi operator-(mi a, const mi& b) { return a -= b; }
    friend mi operator*(mi a, const mi& b) { return a *= b; }
    friend mi operator/(mi a, const mi& b) { return a /= b; }
};


int n;
int c[N], b[N];
int lb[N], psum[N];
mi dp[N][N * N];

mt19937 rng(2333);

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n;
    rep(i, 1, n + 1) cin >> c[i];
    rep(i, 1, n) cin >> b[i];
    int q;
    cin >> q;
    int x; cin >> x;
    lb[1] = psum[1] = x;
    rep(i, 2, n + 1) {
        lb[i] = lb[i - 1] + b[i - 1];
        psum[i] = lb[i] + psum[i - 1];
    }
    dp[0][0] = 1;
    int maxs = 100 * n;
    rep(i, 1, n + 1) {
        rep(s, psum[i],  maxs + 1) {
            rep(cur, 0, min(s + 1, c[i] + 1)) {
                dp[i][s] += dp[i - 1][s - cur];
            }
        }
    }
    mi res = 0;
    rep(i, 0, maxs + 1) res += dp[n][i];
    cout << int(res) << endl;
}