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 operator as op
from functools import reduce

def ncr(n, r):
    if n == 1:
        return 1

    r = min(r, n-r)
    numer = reduce(op.mul, range(n, n-r, -1), 1)
    denom = reduce(op.mul, range(1, r+1), 1)
    return numer // denom


n, m, k = map(int, raw_input().split())

ass = map(int, raw_input().split())
bss = map(int, raw_input().split())

la = [0]
lb = [0]

for elem in ass:
    if elem == 1:
        la[-1] += 1
    else:
        if la[-1] != 0:
            la.append(0)

if la[-1] == 0:
    la.pop(-1)


for elem in bss:
    if elem == 1:
        lb[-1] += 1
    else:
        if lb[-1] != 0:
            lb.append(0)

if lb[-1] == 0:
    lb.pop(-1)

la.sort(key=lambda x: -x)
lb.sort(key=lambda x: -x)


def solve(a, b):
    global la, lb

    result = 0

    for elem in la:
        if elem < a:
            break

        ar = elem-a+1
        for elem2 in lb:
            if elem2 < b:
                break

            br = elem2-b+1
            result += ar*br

    return result


#print la, lb
result = 0
for i in xrange(1, int(k**0.5)+1):
    if k % i == 0:
        now = solve(i, k//i)
        #print i, k//i, now
        result += now

        if i != k//i:
            now = solve(k//i, i)
            #print k//i, i, now
            result += now

print result