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>
#!/usr/bin/env python3
import io, os
from sys import setrecursionlimit
from collections import defaultdict
from itertools import groupby

input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

def rint():
    return map(int, input().split())

def oint():
    return int(input())


def make_freq_line(a):
    freq = defaultdict(int)
    for key, iter in groupby(a):
        if key == 1:
            freq[len(list(iter))] += 1
    return freq

def make_rect(fa, fb):
    rect = defaultdict(int)
    for ka, va in fa.items():
        for kb, vb in fb.items():
            rect[(ka, kb)] += va*vb
    return rect

def find_divisor(n):
    i = 1
    div1 = []
    div2 = []
    while i*i < n:
        if n%i == 0:
            div1.append(i)
            div2.append(n//i)
        i += 1
    if i*i == n:
        div1.append(i)
    return div1 + div2[::-1]


n, m, k = rint()

a = list(rint())
b = list(rint())

fla = make_freq_line(a)
flb = make_freq_line(b)
rect = make_rect(fla, flb)
div = find_divisor(k)

"""
print(n,k,k)
print(dict(fla))
print(dict(flb))
print(dict(rect))
print(div)
"""

ans = 0
for (r, c), v in rect.items():
    for sr in div:
        sc = k//sr
        if r < sr or c < sc:
            continue
        ans += (r-sr+1)*(c-sc+1)*v
print(ans)

