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 collections import defaultdict, Counter
from bisect import bisect, bisect_left
from math import sqrt, gcd, ceil, factorial
from heapq import heapify, heappush, heappop


MOD = 10**9 + 7
inf = float("inf")
ans_ = []

def nin():return int(input())
def ninf():return int(file.readline())

def st():return (input().strip())
def stf():return (file.readline().strip())

def read(): return list(map(int, input().strip().split()))
def readf():return list(map(int, file.readline().strip().split()))

    
ans_ = []


# file = open("input.txt", "r")
def solve():
    n, m, k = read(); 
    a, b = st().replace(" ","").split("0"), st().replace(" ","").split("0")
#     print(a, b)
    a = [len(i) for i in a] 
    b = [len(i) for i in b]
    d1 = defaultdict(int); d2 = defaultdict(int)
    for i in a:
        for j in range(1, i+1):
            d1[j] += i-j+1
    
    for i in b:
        for j in range(1, i+1):
            d2[j] += i-j+1
    
#     print(a, b)
    i = 1; ans = 0
    while i*i <= k:
        if k%i == 0:
#             print(i, k//i, d1, d2)
            ans += d1[i] * d2[k//i] + ((d1[k//i] * d2[i]) if i*i != k else 0)
        i += 1
    ans_.append(ans)
    
        
# file.close()
solve()
for i in ans_:print(i)
    