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 sys import stdin

inf=stdin

n,m,k=map(int,inf.readline().split(" "))
A=list(map(int,inf.readline().split(" ")))
B=list(map(int,inf.readline().split(" ")))

temp1=[0]*(n+1)
temp2=[0]*(m+1)

i=0
while(i<n):
	if(A[i]==0):
		i+=1
		continue
	j=i
	while(j<n and A[j]==1):
		j+=1

	for q in range(1,j-i+1):
		temp1[q]+=j-i-q+1
	i=j

i=0
while(i<m):
	if(B[i]==0):
		i+=1
		continue
	j=i
	while(j<m and B[j]==1):
		j+=1

	for q in range(1,j-i+1):
		temp2[q]+=j-i-q+1
	i=j
ans=0


for i in range(1,len(temp1)):
	if(k%i==0 and k//i<=m):
		ans+=temp1[i]*temp2[k//i]
print(ans)

# for i in dicti:
# 	temp=0
# 	for j in helper:
# 		if(j[0]<=i[0] and j[1]<=i[1]):
# 			temp+=(i[0]-j[0]+1)*(i[1]-j[1]+1)
# 		if(j[1]<=i[0] and j[0]<=i[1] and j[0]!=j[1]):
# 			temp+=(i[0]-j[1]+1)*(i[1]-j[0]+1)
# 	count+=temp*dicti[i]



# def all_partitions(string):
#     for cutpoints in range(1 << (len(string)-1)):
#         result = []
#         lastcut = 0
#         for i in range(len(string)-1):
#             if (1<<i) & cutpoints != 0:
#                 result.append(string[lastcut:(i+1)])
#                 lastcut = i+1
#         result.append(string[lastcut:])
#         yield result

# maxyet=0
# store={'h': -6, 'e': -7, 'l': -8, 'o': 3, 'he': 3, 'hel': -3, 'el': 0, 'hell': 6, 'ell': -5, 'll': 10, 'hello': -3, 'ello': -8, 'llo': 9, 'lo': -6}

# def quality(stri):
# 	return store[stri]

# for partition in all_partitions("hel"):
#     temp=0
#     for i in partition:
#     	temp+=quality(i)
#     if(temp>maxyet):
#     	print(partition)
#     	maxyet=temp

# print(maxyet)