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>
def isqrt(k):
	ret = 0
	x = 2**16
	while x > 0:
		if (ret+x)**2 <= k:
			ret += x
		x //= 2
	return ret


def compress(l):
	n = len(l)
	l.append(0)
	ret = []
	x = y = 0
	i = 0
	while True:
		while l[i] == 0 and i < n:
			i += 1
		if i == n:
			break
		x = i
		while l[i] == 1:
			i += 1
		y = i-1
		ret.append((x, y))
	l.pop()
	return ret

def max_len(a):
	if len(a) == 0:
		return 0
	return max([y-x+1 for (x, y) in a])

def one_list_count(a, n):
	return sum([y-x-n+2 for (x, y) in a if y-x+1 >= n])

def one_div_count(a, b, k, d):
	return one_list_count(a, d) * one_list_count(b, k//d)

def div_less_than_sqrt_count(a, b, k):
	ma = max_len(a)
	mb = max_len(b)
	if mb == 0:
		return 0
	
	ret = 0
	d = (k-1)//mb + 1
	
	while d <= ma and d ** 2 < k:
		if k%d == 0:
			ret += one_div_count(a, b, k, d)
		d += 1
	return ret

n, m, k = tuple(map(int, input().split(' ')))
a = list(map(int, input().split(' ')))
b = list(map(int, input().split(' ')))
a = compress(a)
b = compress(b)

ans = div_less_than_sqrt_count(a, b, k)
ans += div_less_than_sqrt_count(b, a, k)
sk = isqrt(k)
if sk ** 2 == k:
	ans += one_div_count(a, b, k, sk)

print(ans)