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>
import bisect

n=[0]*2
n[0],n[1],k=map(int,input().split())
#a=[]
#b=[]
A=[[],[]]
for i in range(1,401*100):
    if i>k//i:
        break
    if k%i==0:
        if i==k//i:
            A[0].append(i)
            A[1].append(i)
        else:
            A[0]+=[i,k//i]
            A[1]+=[k//i,i]
ans=[1]*len(A[0])
for i in range(2):
    p=map(int,input().split())
    p=list(p)+[0]
    SUM=0
    arr=[]
    for x in p:
        if x==0 and SUM!=0:
            arr.append(SUM)
            SUM=0
        elif x==1:
            SUM+=1
    arr=sorted(arr)
    #a.append(arr)
    
    SUM=sum(arr)
    b=[SUM]
    for x in arr:
        SUM-=x
        b.append(SUM)
    for j,x in enumerate(A[i]):
        place=bisect.bisect_left(arr,x)
        
        ans[j]*=(b[place]-(len(arr)-place)*(x-1))
    
print(sum(ans))