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.
from __future__ import division,print_function
from heapq import*
import sys
le = sys.__stdin__.read().split("\n")[::-1]
mo=10**9+7
if 1:
    n=int(le.pop())
    c = list(map(int,le.pop().split()))
    b = list(map(int,le.pop().split()))
    le.pop()
    x=int(le.pop())
    prefb=[b[0]]
    for k in b[1:]:
        prefb.append(k+prefb[-1])
    prefb.append(0)    #careful when going to C
    prefbt=[0]
    for k in range(1,n-1):
        prefbt.append(k*b[k]+prefbt[-1])
    prefbt.append(0)
sc=sum(c)
d=[[0]*(sc+1) for k in range(n+1)]#fuck a lot of prefix sum in this problem
ds=[[0]*(sc+2) for k in range(n+1)]#prefix array beginning with 0
ds[-1]=list(range(sc+2))
d[-1]=[1]*(sc+1)
for index in range(n-1,-1,-1):
    #can even go linearly for minpref
    minpref=0
    while (minpref-index*prefb[index-1]+prefbt[index-1])/(index+1)<x:
       minpref+=1
    for pref in range(sc+1):
        mi=min(pref+c[index]+1,sc+1)
        ma=max(minpref,pref)
        d[index][pref]=0 if mi<ma else ds[index+1][mi]-ds[index+1][ma]
    for pref in range(1,sc+2):
        ds[index][pref]=(ds[index][pref-1]+d[index][pref-1])%mo
print(d[0][0]%mo)
