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
from math import sqrt
def two_factors(k):
    factors=[]
    for i in range(1, round(sqrt(k))+1):
        if k%i == 0:
            factors.append(i)
            if i*i!=k:
                factors.append(k/i)
    return factors

numbers=list(map(int,input().split()))
n=numbers[0]
m=numbers[1]
k=numbers[2]
a=list(map(int,input().split()))
b=list(map(int,input().split()))

if (k>n*m):
    print(0)
    sys.exit()


counter=0
new_a=[]
new_b=[]
for i in range(0, n):
    if a[i] == 1:
        counter+=1
        if i == n-1:
            new_a.append(counter)
            counter=0
    elif a[i] == 0 and counter!=0 :
        new_a.append(counter)
        counter=0
        
counter=0
for i in range(0, m):
    if b[i] == 1:
        counter+=1
        if i == m-1:
            new_b.append(counter)
            counter=0
    elif b[i] == 0 and counter!=0:
        new_b.append(counter)
        counter=0
total=0
factors=two_factors(k)
for i in factors:
    sum_of_factor_1=0
    sum_of_factor_2=0
    for j in new_a:
        if i<=j:
            factor_1=j-i+1
            sum_of_factor_1+=factor_1
    for x in new_b:
        if k/i<=x:
            factor_2=x-k/i+1
            sum_of_factor_2+=factor_2
    total+=sum_of_factor_1 * sum_of_factor_2

print(round(total))
    