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>

#define mp std::make_pair
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define gc getchar
#define pc putchar

typedef unsigned char u8;
typedef unsigned short u16;
typedef unsigned int u32;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef std::pair <int, int> pii;
typedef std::pair <ll, ll> pll;

const int mod = 1000000007;
const int max_N = 107;

int n, B[max_N], C[max_N], dp[max_N][max_N * max_N];

int read_int() {
	char c = gc(); int ans = 0; bool neg = false;
	while (!isdigit(c)) neg |= (c == '-'), c = gc();
	while (isdigit(c)) ans = 10 * ans + c - '0', c = gc();
	return neg ? -ans : ans;
}

void write_int(int x) {
	if (x < 0) pc('-'), x = -x;
	if (x > 9) write_int(x / 10);
	pc(x % 10 + '0');
}

int min(int x, int y) {return x < y ? x : y;}
int max(int x, int y) {return x > y ? x : y;}
void _min(int &x, int y) {if (x > y) x = y;}
void _max(int &x, int y) {if (x < y) x = y;}

int plus(int x, int y) {return (x += y) >= mod ? x - mod : x;}
int minus(int x, int y) {return (x -= y) < 0 ? x + mod : x;}
int mul(int x, int y) {return (ll)x * y % mod;}
int fma(int x, int y, int z) {return (x + (ll)y * z) % mod;}
int fms(int x, int y, int z) {return (x + (ll)(mod - y) * z) % mod;}
int neg(int x) {return x == 0 ? 0 : mod - x;}
void _plus(int &x, int y) {if ((x += y) >= mod) x -= mod;}
void _minus(int &x, int y) {if ((x -= y) < 0) x += mod;}
void _mul(int &x, int y) {x = (ll)x * y % mod;}
void _fma(int &x, int y, int z) {x = (x + (ll)y * z) % mod;}
void _fms(int &x, int y, int z) {x = (x + (ll)(mod - y) * z) % mod;}
void _neg(int &x) {if (x) x = mod - x;}

int query(int x) {
	dp[0][0] = 1;
	for (int i = 1, sum = 0; i <= n; i++) {
		std::fill(dp[i], dp[i] + sum + C[i] + 1, 0);
		std::partial_sum(dp[i - 1], dp[i - 1] + sum + 1, dp[i - 1], plus);
		for (int j = max(x * i + B[i], 0); j <= sum + C[i]; j++) {
			int l = max(j - C[i], 0), r = min(sum, j);
			if (l <= r) _plus(dp[i][j], minus(dp[i - 1][r], l == 0 ? 0 : dp[i - 1][l - 1]));
		}
		sum += C[i];
	}
	return std::accumulate(dp[n], dp[n] + C[0] + 1, 0ll) % mod;
}

int main() {
	n = read_int();
	for (int i = 1; i <= n; i++) C[i] = read_int();
	C[0] = std::accumulate(C + 1, C + n + 1, 0);
	for (int i = 2; i <= n; i++) B[i] = read_int();
	std::partial_sum(B + 2, B + n + 1, B + 2);
	std::partial_sum(B + 2, B + n + 1, B + 2);
	for (int i = read_int(); i; i--)
		write_int(query(read_int())), pc('\n');
	return 0;
}
