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 lens(arr):
    n = len(arr)
    res = [0]* (n+1)
    i=0
    j=0
    while(i<n):
        while(arr[i]==0):
            i+=1
            if (i>=n):
                break
        j = i
        while(j<n and arr[j]==1):
            j+=1
        for Len in range(1,j-i+1):
            res[Len]+=j-i-Len+1;
        i=j
    return res

t = [int(x) for x in input().split(' ')]
n = t[0]
m = t[1]
k = t[2]
N_arr = [int(x) for x in input().split()]
M_arr = [int(x) for x in input().split()]
#find big rectangle with all 1s
N_lens = lens(N_arr)
M_lens = lens(M_arr)
i=0
j=0
res = 0
i =1
for i in range(1,n+1):
    if(k%i==0 and  k//i<=m):
        res+= N_lens[i] *M_lens[k//i]
    i+=1
print(res)
