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 __future__ import division, print_function
import sys
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

import os, sys, bisect, copy
from collections import defaultdict, Counter, deque
#from functools import lru_cache   #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------
#always use bit 1-indexed


n,m,k = mapi()
a = list(mapi())
b = list(mapi())
res=0
bit = [0]*(n+1)
nit = [0]*(m+1)

def update(bit,n,idx,val):
    while idx<=n:
        bit[idx]+=val
        idx+=idx&(-idx)

def query(bit,idx):
    res = 0
    while idx>0:
        res+=bit[idx]
        idx-=idx&(-idx)
    return res
cnt =0
for i in a:
    if i==1:
        cnt+=1
        update(bit,n,1,1)
        update(bit,n,cnt+1,-1)
    else:
        cnt = 0
cnt =0
for i in b:
    if i==1:
        cnt+=1
        update(nit,m,1,1)
        update(nit,m,cnt+1,-1)
    else:
        cnt = 0
for i in range(1,int(k**.5)+1):
    if k%i!=0: continue
    if i<=n and k//i<=m: 
        res+=query(bit,i)*query(nit,k//i)
    if k//i!=i and k//i<=n and i<=m:
        res+=query(bit,k//i)*query(nit,i)
print(res)
