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>
a,b,k=map(int,input().split())
l=list(map(int,input().split()))
l1=list(map(int,input().split()))
c=0
c1=0
for i in range(a):
    if l[i]==0:
        c+=1
for i in range(b):
    if l1[i]==0:
        c1+=1
f=[0]*(c+1)
g=[0]*(c1+1)
t=0
start=0
l.append(0)
l1.append(0)
for i in range(a+1):
    if l[i]==0:
        f[t]=i-start
        start=i+1
        t+=1
t=0
start=0
for i in range(b+1):
    if l1[i]==0:
        g[t]=i-start
        start=i+1
        t+=1
ans=0
for i in range(1,int(k**0.5)+1):
    if(k%i!=0):
        continue
    n=0
    n1=0
    for j in range(c+1):
        n+=max(0,f[j]-i+1)
    for j in range(c1+1):
        n1+=max(0,g[j]-k//i+1)
    n2=0
    n3=0
    if i==k//i:
        ans+=n*n1
        continue
    for j in range(c+1):
        n2+=max(0,f[j]-k//i+1)
    for j in range(c1+1):
        n3+=max(0,g[j]-i+1)
    ans+=n*n1+n2*n3
print(ans)
    
