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>
# cook your dish here
import math
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
l=[]
for i in range(2,math.floor(math.sqrt(k))+1):
    if k%i==0:
        l.append((i,k//i))
l.append((1,k))
count=0
d={}
r={}
for i in range(n):
    if a[i]==1:
        count+=1
    else:
        if count!=0:
            d[count]=d.get(count,0)+1
            count=0
if count!=0:
    d[count]=d.get(count,0)+1
    count=0
for i in range(m):
    if b[i]==1:
        count+=1
    else:
        if count!=0:
            r[count]=r.get(count,0)+1
            count=0
if count!=0:
    r[count]=r.get(count,0)+1
    count=0
ans=0
#print(d,r,l)
for i in l:
    ans1,ans2,ans3,ans4=0,0,0,0
 
    for j in d:
        if i[0]<=j:
            ans1=ans1+d[j]*(j-i[0]+1)
        if i[1]<=j:
            ans2=ans2+d[j]*(j-i[1]+1)
    for j in r:
        if i[0]<=j:
            ans3=ans3+r[j]*(j-i[0]+1)
        if i[1]<=j:
            ans4=ans4+r[j]*(j-i[1]+1)
  #  print(ans1,ans2,ans3,ans4)
    if i[0]==i[1]:
        ans+=ans1*ans3
    else:
        ans=ans+ans1*ans4+ans2*ans3
    #print(ans)
print(ans)