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>
def make_divisors(n):
    divisors = []
    for i in range(1, int(n ** 0.5) + 1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n // i)
    # divisors.sort()
    return divisors

import itertools
def countstrs(s):
    return [(k, len(list(g))) for k, g in itertools.groupby(s)]

from pprint import pprint
n,m,k = map(int, input().split())
data = input().split()
datb = input().split()

divs = make_divisors(k)
rects = []
for x in divs:
    rects.append((x, k // x))
#print("rects")
#print(rects)

from collections import Counter, defaultdict
import sys

da = defaultdict(int)
db = defaultdict(int)

cnt = 0
l = len(data)
for i in range(l):
    if data[i] == "1":
        cnt += 1
    if data[i] == "0" or i == l - 1:
        da[cnt] += 1
        cnt = 0
#print(da)

cnt = 0
l = len(datb)
for i in range(l):
    if datb[i] == "1":
        cnt += 1
    if datb[i] == "0" or i == l - 1:
        db[cnt] += 1
        cnt = 0
#print(db)

res = 0
for lx in da:
    for ly in db:
        for x, y in rects:
            reccount = da[lx] * db[ly]
            #print("{0} x {1} count = {4}, find {2} x {3} rect".format(lx, ly, x, y, reccount))
            cx = lx - x
            cy = ly - y
            if cx > -1 and cy > -1:
                cannum = (cx + 1) * (cy + 1) * reccount
                res += cannum
print(res)