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,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
from bisect import bisect_left
cna=[]
cnt=0
i=0
while i<n:
	if a[i]==1:
		cnt+=1
	else:
		cna.append(cnt)
		cnt=0
	i+=1
if a[-1]==1:cna.append(cnt)
cna.sort()
ca=[0]
for i in range(len(cna)):
	ca.append(ca[-1]+cna[i])

cnb=[]
cnt=0
i=0
while i<m:
	if b[i]==1:
		cnt+=1
	else:
		cnb.append(cnt)
		cnt=0
	i+=1
if b[-1]==1:cnb.append(cnt)
cnb.sort()

cb=[0]
for i in range(len(cnb)):
	cb.append(cb[-1]+cnb[i])

pd=[]
for i in range(1,int(k**0.5)+1):
	if k%i==0:
		if i==k//i and i<=n and i<=m: pd.append([i,i])
		else:
			if i<=n and k//i<=m:
				pd.append([i,k//i])
			if k//i<=n and i<=m:
				pd.append([k//i,i])
ans=0
for f,g in pd:
	pa=bisect_left(cna,f)
	pb=bisect_left(cnb,g)
	sa=ca[-1]-ca[pa]
	sb=cb[-1]-cb[pb]
	na=len(cna)-pa
	nb=len(cnb)-pb
	ans+=(sa-na*f+na)*(sb-nb*g+nb)
print(ans)
