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
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__

def divisor(n):
    ret = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            ret.append(i)
            if i*i != n:
                ret.append(n//i)
    return ret

class Binary_Indexed_Tree_RAQ():
    def __init__(self, n):
        self.data = [0]*(n+1)
        self.n = n+1

    def __add(self, ind, num):
        while(ind <= self.n):
            self.data[ind - 1] += num
            ind += ind & -ind

    def range_add(self, s, t, x):
        self.__add(s,x)
        self.__add(t+1,-x)

    def query(self, i):
        ret = 0
        while(i > 0):
            ret += self.data[i-1]
            i -= i & -i
        return ret

n,m,k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))

cnt = 0
cnt_a = Binary_Indexed_Tree_RAQ(n)
cnt_b = Binary_Indexed_Tree_RAQ(m)
for i in a:
    if i == 1:
        cnt += 1
        cnt_a.range_add(1,cnt, 1)
    else:
        cnt = 0

cnt = 0
for i in b:
    if i == 1:
        cnt += 1
        cnt_b.range_add(1,cnt, 1)
    else:
        cnt = 0

ans = 0
for i in divisor(k):
    j = k//i
    if i <= n and j <= m:
        ans += cnt_a.query(i)*cnt_b.query(j)

print(ans)

