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
def countDivisors(n) : 
    l={}
    for i in range(1, int((math.sqrt(n)) + 1)) : 
        if (n % i == 0) : 
              
            # If divisors are equal, 
            # count only one 
            if (n / i == i) : 
                l[i]=0
            else : # Otherwise count both 
                l[i]=0
                l[n//i]=0
    return l
n,m,k=map(int,input().split())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
j=0
c1,c2=0,0
f1=0
f2=0
a1=[0 for i in range(n+1)]
a2=[0 for i in range(m+1)]
while j<max(n,m):
    if f1<n:
        if l1[j]==1:
            c1+=1
        else:
            if c1!=0:
                a1[c1]+=1
            c1=0
    if f2<m:
        if l2[j]==1:
            c2+=1
        else:
            if c2!=0:
                a2[c2]+=1
            c2=0
    j+=1
    f1+=1
    f2+=1
if c1!=0:
    a1[c1]+=1
if c2!=0:
    a2[c2]+=1

l3=countDivisors(k)
l4=l3.copy()
ans=[0 for i in range(n)]
for i in range(max(m+1,n+1)):
    if i<n+1 and a1[i]>0:
        for j in sorted(l3.keys()):
            if i>=j:
                l3[j]+=(a1[i]*(i-j+1))
                continue
            break
    if i<m+1 and a2[i]>0:
        for j in sorted(l4.keys()):
            if i>=j:
                l4[j]+=(a2[i]*(i-j+1))
                continue
            break
c=0
'''print(l3,l4)
print(a1,a2)'''
for i in l3:
    c+=(l3[i]*l4[k//i])
print(c)
    
        

        
    