You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.

How many subrectangles of size (area) k consisting only of ones are there in c?

A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.

The size (area) of a subrectangle is the total number of cells in it.

Input

The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.

The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.

Output

Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.

Examples

Input


3 3 2
1 0 1
1 1 1


Output


4


Input


3 5 4
1 1 1
1 1 1 1 1


Output


14

Note

In first example matrix c is:

<image>

There are 4 subrectangles of size 2 consisting of only ones in it:

<image>

In second example matrix c is:

<image>
from math import sqrt

n,m,k = map(int,input().split())

l1 = list(map(int,input().split()))
l2 = list(map(int,input().split()))

r = []
s = 0
for i in range(len(l1)):

	if i > 0 and l1[i] == l1[i-1] and l1[i] == 1:
		s = s + 1
	elif l1[i] == 0:
		r.append(s)
		s = 0
	elif l1[i] == 1:
		s = 1
r.append(s)
r.sort()
r1 = []
s = 0
for i in range(len(l2)):

	if i >0 and l2[i] == l2[i-1] and l2[i] == 1:
		s = s + 1
	elif l2[i] == 0:
		r1.append(s)
		s = 0
	elif l2[i] == 1:
		s = 1

r1.append(s)
r1.sort()

fact = []
for i in range(1,int(sqrt(k))+1):

	if k%i == 0 and k%(k//i) == 0 and i != k//i:
		fact.append(i)
		fact.append(k//i)
	elif k//i == i and k%i == 0:
		fact.append(i)

fact.sort()


d1 = {fact[i]:0 for i in range(len(fact))}
d2 = {fact[i]:0 for i in range(len(fact))}

store1 = [0 for i in range(len(r)+1)]

for i in range(1,len(r)+1):
	store1[i] = store1[i-1] + r[i-1]

i = 0
j = 0 

# print(r,r1)
x = len(store1)-1
while i <= len(fact)-1 and j <= len(r)-1:
	if fact[i] <= r[j]:
		d1[fact[i]] = d1[fact[i]] + store1[x]-store1[j]-(x-j)*fact[i] + x-j
		i=i+1
	elif fact[i] > r[j]:
		j = j + 1


store2 = [0 for i in range(len(r1)+1)]
y = len(store2)-1
for i in range(1,len(r1)+1):
	store2[i] = store2[i-1]+r1[i-1]
i = 0
j = 0
while i <= len(fact)-1 and j <= len(r1)-1:

	if fact[i] <= r1[j]:
		
		d2[fact[i]] = d2[fact[i]] + store2[y]-store2[j]-(y-j)*fact[i] + y-j
		i = i + 1
	elif fact[i] > r1[j]:
		j = j + 1

ans = 0
for i in d1:
	ans = ans + d1[i]*d2[k//i]
print(ans)