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>
N,M,K=map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
pA=[0]*(N+1)
for i in range(N):
    pA[i+1]=pA[i]+A[i]
pB=[0]*(M+1)
for i in range(M):
    pB[i+1]=pB[i]+B[i]
ans=0
for p in range(1,int(K**0.5)+1):
    if K%p>0: continue
    q=K//p
    if p>q: break

    X=0
    for i in range(N+1-p):
        if pA[i+p]-pA[i]==p: X+=1
    Y=0
    for i in range(M+1-q):
        if pB[i+q]-pB[i]==q: Y+=1
    ans+=X*Y
    if p==q: break
    X=0
    for i in range(N+1-q):
        if pA[i+q]-pA[i]==q: X+=1
    Y=0
    for i in range(M+1-p):
        if pB[i+p]-pB[i]==p: Y+=1
    ans+=X*Y
print(ans)