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>
import sys,heapq
input = sys.stdin.readline

def One(type):
	return (type(input()))
def List(type):
	return list(map(type,input().split()))
def Str():
	return (list(input()))
def Many(type):
	return map(type,input().split())

def calc(a):
	i = 0
	n = len(a)
	ans  = [0]*n
	while i<n:
		if a[i]==0: 
			i+=1
			continue

		j = i 
		while j<n and a[j]==1:
			j+=1

		for cur in range(j-i):
			ans[cur]+=j-i-cur

		i = j+1

	return ans

def main():
	n,m,k = Many(int)
	a,b = List(int),List(int)

	c1,c2 = calc(a),calc(b)
	ans = 0


	for i,nm in enumerate(c1):
		if nm==0: continue;

		if k/(i+1)<=m and k%(i+1)==0:
			num = int(k/(i+1))
			ans+=nm*c2[num-1]

	print(ans)

main()
