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 math
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
a_size,b_size=[],[]
temp=0
for i in range(len(a)):
    if(a[i]==1):
        temp+=1
    if(a[i]==0):
        a_size.append(temp)
        temp=0
if(temp!=0):a_size.append(temp)
temp=0
for i in range(len(b)):
    if(b[i]==1):
        temp+=1
    if(b[i]==0):
        b_size.append(temp)
        temp=0
if(temp!=0):b_size.append(temp)
#a->row b->col
'''row,col=dict(),dict()
x,y=set(),set()
for i in a_size:
    if(i in x):
        row[i]+=1
    else:
        row[i]=1
for i in b_size:
    if(i in y):
        col[i]+=1
    else:
        col[i]=1
'''
ans=0
rr=int(int((k)**0.5)+1)
for i in range(1,rr):
    if(k%i==0):
        n1=k/i
        n2=i
        temp1,temp2=0,0
        for val in a_size:
            if val>=n1:
                temp1+=int(val-n1+1)
        for val in b_size:
            if val>=n2:
                temp2+=int(val-n2+1)
        ans+=int(temp1*temp2)
        #print(temp1,temp2,ans)
        if(n1!=n2):
            n1,n2=n2,n1
            temp1,temp2=0,0
            for val in a_size:
                if val>=n1:
                    temp1+=int(val-n1+1)
            for val in b_size:
                if val>=n2:
                    temp2+=int(val-n2+1)
            ans+=temp1*temp2
print(ans)

