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())
a = map(int, input().split())
b = map(int, input().split())

def make(a):
	d = dict()
	s = 0
	for v in a:
		s += v
		if v == 0 and s != 0:
			if s not in d:
				d[s] = 0
			d[s] += 1
			s = 0
	if s != 0:
		if s not in d:
			d[s] = 0
		d[s] += 1
	return d

def count(i, j, r, c):
	cnt = 0
	for _r, x in r:
		if i > _r:
			break
		for _c, y in c:
			if j > _c:
				break
			cnt += (_r - i + 1) * (_c - j + 1) * x * y
	return cnt

r = make(a)
c = make(b)
r = sorted(r.items(), key=lambda x: -x[0])
c = sorted(c.items(), key=lambda x: -x[0])
ans = 0
l = int(sqrt(k) + 1e-4) + 1
for i in range(1, l):
	if (k % i) == 0:
		j = k // i
		ans += count(i, j, r, c)
		if i != j:
			ans += count(j, i, r, c)
	
print(ans)
