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
read = lambda: list(map(int, sys.stdin.readline().strip().split()))

n, m , k = read()
a = read()
b = read()
ca = [0]*(n+1)
cb = [0]*(m+1)
countb = 0
for num in b:
    if num == 1:
        countb += 1
    else:
        if countb != 0:
            cb[countb] += 1
            countb = 0
cb[countb] += 1


counta = 0
for num in a:
    if num == 1:
        counta += 1
    else:
        ca[counta] += 1
        counta = 0
ca[counta] += 1
counta = 0
#print(ca)
if n-1 >= 1:
    ca[n-1] = 2*ca[n]
for i in range(n-2, 0, -1):
    counta = ca[i+2]
    ca[i] += 2*ca[i+1]
    ca[i] -= counta
    #print(ca)

countb = 0
if m-1 >= 1:
    cb[m-1] = 2*cb[m]
for i in range(m-2, 0, -1):
    countb = cb[i+2]
    cb[i] += 2*cb[i+1]
    cb[i] -= countb
    #print(cb)


#print(ca)
#print(cb)
ans = 0
for i in range(1, n+1):
    if ca[i] > 0 and k % i==0 and k//i <= m:
        ans += (ca[i]*cb[k//i])

print(ans)









