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>
n, m, k = map(int, raw_input().split())
a = map(int, raw_input().split()) + [0]
b = map(int, raw_input().split()) + [0]

ca = [0 for i in range(n+10)]
c = 0
for i in range(n+1):
	if a[i] == 1:
		c += 1
	else:
		ca[1] += c
		ca[2] -= c+1
		ca[c+2] += 1
		c = 0
# print ca
for j in range(2):
	x = 0
	for i in range(n+10):
		x += ca[i]
		ca[i] = x
	# print ca

cb = [0 for i in range(m+10)]
c = 0
for i in range(m+1):
	if b[i] == 1:
		c += 1
	else:
		cb[1] += c
		cb[2] -= c+1
		cb[c+2] += 1
		c = 0
# print cb
for j in range(2):
	x = 0
	for i in range(m+10):
		x += cb[i]
		cb[i] = x
	# print cb

ans = 0
d = 1
while d*d <= k:
	if d*d == k:
		if d<n+10 and k/d<m+10:
			ans += ca[d]*cb[k/d]
	elif k%d == 0:
		if d<n+10 and k/d<m+10:
			ans += ca[d]*cb[k/d]
		if k/d<n+10 and d<m+10:
			ans += ca[k/d]*cb[d]
	d += 1
print ans