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>
input = raw_input

n, m, k = map(int, input().split(" "))
a_vec = map(int, input().split(" "))
b_vec = map(int, input().split(" "))

nums = []
a_renzokus = [] 
n_renzoku = 0
for i in range(n):
    a = a_vec[i]
    if a == 0:
        n_renzoku = 0
    else:
        n_renzoku += 1
    a_renzokus.append(n_renzoku)

max_a_renzoku = max(a_renzokus)

b_renzokus = []
n_renzoku = 0
for j in range(m):
    b = b_vec[j]
    if b == 0:
        n_renzoku = 0
    else:
        n_renzoku += 1
    b_renzokus.append(n_renzoku)
max_b_renzoku = max(b_renzokus)

def get_div_pairs(n):
    pairs = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            #divisors.append(i)
            d1 = i
            d2 = n // i
            pairs.append([d1, d2])
            
    return pairs

div_pairs = get_div_pairs(k)
#print(yoko_maxs, tate_maxs)

out = 0
for x1, x2 in div_pairs:
    if x1 != x2:
        cur_pairs = [[x1, x2], [x2, x1]]
    else:
        cur_pairs = [[x1, x2]]
    for d1, d2 in cur_pairs:
        if max_a_renzoku < d1 or max_b_renzoku < d2:
            continue

        ok_num_a = 0 
        for i in range(n):
            cur_a_renzoku = a_renzokus[i]
            if cur_a_renzoku >= d1:
                ok_num_a += 1
            
        ok_num_b = 0 
        for j in range(m):
            cur_b_renzoku = b_renzokus[j]
            if cur_b_renzoku >= d2:
                ok_num_b += 1
        out += ok_num_a * ok_num_b

print(out)