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


# def fac(n):
#     f = []
#     for i in range(1, int(sqrt(n)) + 1):
#         if n % i == 0:
#             f.append(i)
#     return f


n, m, k = map(int, stdin.readline().split())
a = list(map(int, stdin.readline().split()))
b = list(map(int, stdin.readline().split()))
# mat = [[0 for _ in range(m)] for j in range(n)]
# for i in range(n):
#     for j in range(m):
#         mat[i][j] = a[i] * b[j]
sa,sb=[0]*50000, [0]*50000
i, j = 0, 0
while i<n:
    j=i
    while j<n:
        if a[j]==0:break
        j+=1
    if j>i:
        temp=j-i
        c=1
        for x in range(temp,0,-1):
            sa[x]+=c
            c+=1
    i=j+1
i, j = 0, 0
while i<m:
    j=i
    while j<m:
        if b[j]==0:break
        j+=1
    if j>i:
        temp=j-i
        c=1
        for x in range(temp,0,-1):
            sb[x]+=c
            c+=1
    i=j+1

rect=0
for x in range(1,int(sqrt(k))+1):
    if k%x==0:
        if max(x,k//x)<=max(n,m):
            rect+=sa[x]*sb[k//x]
            if x!=k//x:
                rect+=sa[k//x]*sb[x]
print(rect)





