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 math, collections, sys
input = sys.stdin.readline
factors = set()
def Divisors(n): 
    i = 1
    while i <= math.sqrt(n):
        if (n % i == 0):
            factors.add(i)
            factors.add(n//i)
        i+=1
n, m, k = map(int, input().split())
Divisors(k)
a = list(map(int, input().split()))
b = list(map(int, input().split()))
countA = collections.defaultdict(int)
countB = collections.defaultdict(int)
za, zb = [0], [0]
for i in a:
    za.append(za[-1]+i)
for i in b:
    zb.append(zb[-1]+i)
for i in factors:
    for j in range(n-i+1):
        if za[j+i] - za[j] == i:
            countA[i]+=1
    for j in range(m-i+1):
        if zb[j+i] - zb[j] == i:
            countB[i]+=1
ans = 0
x = 1
for i in factors:
    ans+=countA[i]*countB[k//i]
print(ans)