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.

Difficulty: VERY_HARD
Tags: Dynamic programming

Mathematics
Skill types: Dynamic programming

Starter code:

Solutions:
def putin():
	return map(int, input().split())

def sol():
	n = int(input())
	C = list(putin())
	B = list(putin())
	q = int(input())
	x = int(input())
	min_arr = [x]
	min_part_sums = [x]
	part_sums = [C[0]]
	for i in range(1, n):
		part_sums.append(part_sums[-1] + C[i])
	for elem in B:
		min_arr.append(min_arr[-1] + elem)
		min_part_sums.append(min_arr[-1] + min_part_sums[-1])
	for i in range(n):
		if min_part_sums[i] > part_sums[i]:
			return 0
	if min_part_sums[0] > C[0]:
		return 0
	answer = [1] * (part_sums[0] - max(0, min_part_sums[0]) + 1)
	for k in range(1, n):
		new_answer = [0] * (part_sums[k] - max(0, min_part_sums[k]) + 1)
		cnt = 1
		window = answer[-1]
		new_answer[-1] = window
		while cnt <= len(new_answer) - 1:
			cnt += 1
			if cnt <= len(answer):
				window += answer[-cnt]
			if C[k] + 1 < cnt:
				window -= answer[C[k] + 1 - cnt]
			new_answer[-cnt] = window
		answer = new_answer.copy()
	m = 10 ** 9 + 7
	return sum(answer) % m
print(sol())


import sys
input = lambda : sys.stdin.readline().rstrip()
N = int(input())
C = [int(a) for a in input().split()]
B = [int(a) for a in input().split()]
Q = int(input())
x = int(input())
P = 10 ** 9 + 7
dp = [[0] * 20100 for _ in range(N + 1)]
dp[0][0] = 1
ans = 0
s = x
t = s
for i in range(N):
	for j in range(20050, t - 1, -1):
		if j < 0:
			break
		dp[i + 1][j] = (dp[i + 1][j + 1] + dp[i][max(j - C[i], 0)] - dp[i][j + 1]) % P
	for j in range(min(t - 1, 20050), -1, -1):
		dp[i + 1][j] = dp[i + 1][j + 1]
	if i < N - 1:
		s += B[i]
		t += s
print(dp[-1][0] % P)


def solve():
	MOD = 10 ** 9 + 7
	n = int(input())
	c = list(map(int, input().split()))
	b = [0] + list(map(int, input().split()))
	q = int(input())
	queries = list(map(int, input().split()))
	maxans = 1
	for c1 in c:
		maxans = maxans * (c1 + 1) % MOD
	ans = {}
	for i in range(1, n):
		b[i] += b[i - 1]
	s = lb = 0
	for i in range(1, n):
		s -= b[i]
		lb = min(lb, s // (i + 1))
	s = ub = c[0]
	for i in range(n):
		s += c[i] - b[i]
		ub = min(ub, s // (i + 1))
	for x in queries:
		if x <= lb:
			print(maxans)
		elif x > ub:
			print(0)
		elif x in ans:
			print(ans[x])
		else:
			dp0 = [1] * 10002
			dp0[0] = 0
			bd = 0
			for i in range(n):
				dp1 = [0] * 10002
				bd += b[i] + x
				for j in range(max(bd, 0), 10001):
					dp1[j + 1] = (dp1[j] + dp0[j + 1] - dp0[max(j - c[i], 0)]) % MOD
				dp0 = dp1[:]
			a = dp0[-1]
			ans[x] = a
			print(a)
import sys
input = lambda : sys.stdin.readline().rstrip()
solve()