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 pair(k):
    factor=[]
    x,y=k,1
    for i in range(1,int(k**0.5)+1):
        if(k%i==0 and i<n+1 and k//i<m+1):
            factor.append((i,k//i))
        if(k%i==0 and i!=k//i and k//i<n+1 and i<m+1):
            factor.append((k//i,i))
    
    return factor
    
    
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
comb=[]
if(k==1):
    oneA,oneB=0,0
    for i in a:
        if(i==1):
            oneA=oneA+1
    for i in b:
        if(i==1):
            oneB=oneB+1
    print(oneB*oneA)
else:
    comb=pair(k)
    ans=0
    # print(comb)
    for i in range(len(comb)):
        itr=comb[i][0]
        count1=0;c1=0
        for j in a:
            if(j==1):
                c1=c1+1
            else:
                if(c1>itr or c1==itr):
                    count1=count1+(c1-itr+1)
                c1=0
        if(c1>itr or c1==itr):
            count1=count1+(c1-itr+1)
        c1=0
        itr=comb[i][1]
        count2=0
        for j in b:
            if(j==1):
                c1=c1+1
            else:
                if(c1>itr or c1==itr):
                    count2=count2+(c1-itr+1)
                c1=0
        if(c1>itr or c1==itr):
            count2=count2+(c1-itr+1)
                
        ans=ans+count2*count1        
                
                    
            
            
        
        
    print(ans)