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.
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#endif

#include <bits/stdc++.h>

using namespace std;

typedef long double ld;

#ifdef DEBUG
#define eprintf(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#else
#define eprintf(...) ;
#endif

#define sz(x) ((int) (x).size())
#define TASK "text"

const int inf = (int) 1.01e9;
const long long infll = (long long) 1.01e18;
const ld eps = 1e-9;
const ld pi = acos((ld) -1);

#ifdef DEBUG
mt19937 mrand(300); 
#else
mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count()); 
#endif

int rnd(int x) {
  return mrand() % x;
}

void precalc() {
}

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

int mul(int a, int b) {
  return (long long) a * b % mod;
}

void add(int &a, int b) {
  a += b;
  if (a >= mod) {
    a -= mod;
  }
}

const int maxn = 105;
int n;
int c[maxn], b[maxn];

bool read() {
  if (scanf("%d", &n) < 1) {
    return false;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
  }
  for (int i = 0; i + 1 < n; ++i) {
    scanf("%d", &b[i]);
  }
  return true;
}

const int maxx = maxn * maxn;
int dp[maxx], ndp[maxx];

void solve() {
  int q;
  scanf("%d", &q);
  for (int qq = 0; qq < q; ++qq) {
    int x;
    scanf("%d", &x);
    for (int i = 0; i < maxx; ++i) {
      dp[i] = 0;
    }
    dp[0] = 1;
    int s = 0;
    for (int i = 0; i < n; ++i) {
      s += x;
      int sum = 0;
      for (int j = 0, k = 0; j < maxx; ++j) {
        add(sum, dp[j]);
        while (j - k > c[i]) {
          add(sum, mod - dp[k]);
          ++k;
        }
        ndp[j] = (j < s ? 0 : sum);
      }
      swap(dp, ndp);
      if (i + 1 < n) {
        x += b[i];
      }
    }
    int res = 0;
    for (int i = 0; i < maxx; ++i) {
      add(res, dp[i]);
    }
    printf("%d\n", res);
  }
}

int main() {
  precalc();
#ifdef DEBUG
  assert(freopen(TASK ".in", "r", stdin));
  assert(freopen(TASK ".out", "w", stdout));
#endif
  while (read()) {
    solve();
#ifdef DEBUG
    eprintf("Time %.2f\n", (double) clock() / CLOCKS_PER_SEC);
#endif
  }
  return 0;
}
