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
import os
import time
import collections
from collections import Counter, deque
import itertools
import math
import timeit
import random
import string

#########################
# imgur.com/Pkt7iIf.png #
#########################

def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r

def divs(n, start=1):
    divisors = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors

def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t

def flin(d, x, default=-1):
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return default, default
    else:
        return left, right

def ceil(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def prr(a, sep=' '): print(sep.join(map(str, a)))
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)


# input = sys.stdin.readline

n, m, k = mi()
a = li()
b = li()
d = [[] for _ in range(n)]

# for i in range(n):
#     for j in range(m):
#         d[i].append(a[i] * b[j])
#
# for row in d:
#     print(row)

ac = dd()
bc = dd()
i = 0
while i < n:
    c = 0
    while i < n and a[i] == 1:
        c += 1
        i += 1
    if c > 0: ac[c] += 1
    i += 1
i = 0
while i < m:
    c = 0
    while i < m and b[i] == 1:
        c += 1
        i += 1
    if c > 0: bc[c] += 1
    i += 1

r = {}
for i in ac.keys():
    for j in bc.keys():
        r[(i, j)] = ac[i] * bc[j]

d = divs(k, 1)
res = 0
for (i, j) in r:
    m = r[(i,j)]
    for q in d:
        x = max(0, i - q + 1)
        y = max(0, j - (k//q) + 1)
        res += m*x*y
print(res)








