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>
def countseg(a,b,x,y):
    #print(x,y)
    seg1=0
    seg2=0
    for j in range(len(a)):
        if a[j]>=x:
            seg1=seg1+a[j]-x+1
    for j in range(len(b)):
        if b[j]>=y:
            seg2=seg2+b[j]-y+1
    #print(seg1,seg2)
    return (seg1*seg2)
from math import sqrt
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
ma=list()
mb=list()
c=0
for i in range(n):
    if a[i]==1:
        c=c+1
    else:
        ma.append(c)
        c=0
if a[n-1]==1:
    ma.append(c)
#print(ma)
c=0
for i in range(m):
    if b[i]==1:
        c=c+1
    else:
        mb.append(c)
        c=0
if b[m-1]==1:
    mb.append(c)
#print(mb)
ans=0
for i in range(1,int(sqrt(k))+1):
    if k%i==0:
        x=i
        y=k//i
        ans = ans+ countseg(ma,mb,x,y)
        if x!=y:
            ans=ans+ countseg(ma,mb,y,x)
print(ans)
        
    
