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 bisect import bisect_left as bl, bisect_right as br, insort
import sys
import heapq
from math import *
from collections import defaultdict as dd, deque
def data(): return sys.stdin.readline().strip()
def mdata(): return map(int, data().split())
#sys.setrecursionlimit(100000)

def fac(x):
    f=set()
    for i in range(1,int(x**0.5)+1):
        if x%i==0:
            f.add(i)
            f.add(x//i)
    return f

n,m,k=mdata()
a=list(mdata())
b=list(mdata())
f=sorted(fac(k))
t=0
a1={}
b1={}
fa={}
fb={}
for i in a:
    if i==1:
        t+=1
    elif t!=0:
        if a1.get(t)==None:
            a1[t]=1
        else:
            a1[t]+=1
        t=0
if t!=0:
    if a1.get(t) == None:
        a1[t] = 1
    else:
        a1[t] += 1
for i in f:
    fa[i]=0
    for j in a1:
        if i<=j:
            fa[i]+=(j-i+1)*a1[j]

t=0
for i in b:
    if i==1:
        t+=1
    elif t!=0:
        if b1.get(t)==None:
            b1[t]=1
        else:
            b1[t]+=1
        t=0
if t!=0:
    if b1.get(t) == None:
        b1[t] = 1
    else:
        b1[t] += 1
for i in f:
    fb[i] = 0
    for j in b1:
        if i<=j:
            fb[i]+=(j-i+1)*b1[j]

ans=0
for i in fa:
    if k//i in fb:
        ans+=fa[i]*fb[k//i]
print(ans)