1323_B. Count Subrectangles
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>
```python
n, m, k = map(int, input().split())           
                                              
def intervals(arr):                           
    res = []                                  
    curr = 0                                  
    for e in arr:                             
        if e == 1:                            
            curr += 1                         
        if e == 0:                            
            if curr != 0:                     
                res.append(curr)              
                curr = 0                      
    if curr != 0:                             
        res.append(curr)                      
    return res                                
                                              
a = intervals(list(map(int, input().split())))
b = intervals(list(map(int, input().split())))
                                              
factors = []                                  
for x in range(1, int(k**(1/2))+1):           
    if k % x == 0:                            
        factors.append(x)                     
                                              
def count(ivs, x):                            
    total = 0                                 
    for iv in ivs:                            
        total += max(iv-x+1, 0)               
    return total                              
                                              
total = 0                                     
for f in factors:                             
    one = count(a, f) * count(b, k//f)        
    if f == k//f:                             
        total += one                          
        continue                              
    two = count(a, k//f) * count(b, f)        
    total += one                              
    total += two                              
                                              
print(total)                                  
```