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 = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
pairs = []
for num1 in range(1, int(((k)**(1/2)))+1):
    num2 = k/num1
    if num2.is_integer():
        pairs.append((num1, int(num2)))
contig_a = []
contig_b = []
ongoing = 0
#print(pairs)
for num in a:
    if num == 1:
        ongoing += 1
    elif num == 0:
        if ongoing > 0:
            contig_a.append(ongoing)
        ongoing = 0
if ongoing > 0:
    contig_a.append(ongoing)
    ongoing = 0
for num in b:
    if num == 1:
        ongoing += 1
    elif num == 0:
        if ongoing > 0:
            contig_b.append(ongoing)
        ongoing = 0
if ongoing > 0:
    contig_b.append(ongoing)
    ongoing = 0
total = 0
#print(contig_a, contig_b)
for pair in pairs:
    c, d = pair
    firsta = [num for num in contig_a if num >= c]
    firstb = [num for num in contig_b if num >= d]
    #print(firsta, firstb)
    for x in firsta:
        for y in firstb:
            total += (x-c+1)*(y-d+1)
    if c != d:
        firsta = [num for num in contig_a if num >= d]
        firstb = [num for num in contig_b if num >= c]
        #print(firsta, firstb)
        #print(d, c)
        for x in firsta:
            for y in firstb:
                total += (x-d+1)*(y-c+1)
        #print(firsta, firstb)
print(total)
