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
input = sys.stdin.readline
def fastio():
    from io import StringIO
    from atexit import register
    global input
    sys.stdin = StringIO(sys.stdin.read())
    input = lambda : sys.stdin.readline().rstrip('\r\n')
    sys.stdout = StringIO()
    register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()

INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br

n, m, k = I()
a = I()
b = I()

ans = 0
i, j = 0, 0
start_a, start_b = 0, 0

def factors(x):
    res = []
    for i in range(1, int(x ** 0.5) + 1):
        if x % i: continue
        res.append(i)
    return res

f = factors(k)
x, y = dd(int), dd(int)

cur = 0
for i in range(n):
    cur += a[i]
    if a[i] == 0:
        x[cur] += 1
        cur = 0
if cur:
    x[cur] += 1
cur = 0
for i in range(m):
    cur += b[i]
    if b[i] == 0:
        y[cur] += 1
        cur = 0
if cur:
    y[cur] += 1

for len_a in x:
    count_a = x[len_a]
    for len_b in y:
        count_b = y[len_b]
        for fac in f:
            a, b = fac, k // fac
            if len_a >= a and len_b >= b:
                ans += max(0, (len_a - a + 1) * (len_b - b + 1) * count_a * count_b)
            if len_a >= b and len_b >= a and a != b:
                ans += max(0, (len_a - b + 1) * (len_b - a + 1) * count_b * count_a)
print(ans)