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>
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
 
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
 
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]

n,m,k = li()
a = li()
b = li()
d = defaultdict(int)
tot = 0
last = a[0]
curr = a[0]
for i in range(1,n):
    if a[i] == 1:
        if last == 1:
            curr += 1
        else:
            curr = 1
            last = 1
    else:
        add = 1
        for j in range(curr,0,-1):
            d[j] += add
            add += 1
        # d[curr] += 1
        curr = 0
        last = 0
if curr:
    add = 1
    for j in range(curr,0,-1):
        d[j] += add
        add += 1

d2 = defaultdict(int)
curr = last = b[0]
for i in range(1,m):
    if b[i] == 1:
        if last == 1:
            curr += 1
        else:
            curr = 1
            last = 1
    else:
        add = 1
        for j in range(curr,0,-1):
            d2[j] += add
            add += 1
        curr = 0
        last = 0
if curr:
    add = 1
    for j in range(curr,0,-1):
        d2[j] += add
        add += 1
tot = 0

# print(d,d2,k)
for i in factors(k):
    tot += d[i]*d2[k//i]
print(tot)