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 sys
import math
from collections import defaultdict,Counter

# input=sys.stdin.readline
# def print(x):
#     sys.stdout.write(str(x)+"\n")

# sys.stdout=open("CP1/output.txt",'w')
# sys.stdin=open("CP1/input.txt",'r')

# m=pow(10,9)+7
n,m,k1=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=0
l1=[0]*(n+1)
l2=[0]*(m+1)
c1=0
c2=0
for j in range(n):
    if a[j]:
        c+=1
    else:
        for k in range(1,c+1):
            l1[k]+=c-k+1
        c=0
    c1=max(c,c1)
for k in range(1,c+1):
    l1[k]+=c-k+1
c=0
for j in range(m):
    if b[j]:
        c+=1
    else:
        for k in range(1,c+1):
            l2[k]+=c-k+1
        c=0
    c2=max(c,c2)
for k in range(1,c+1):
    l2[k]+=c-k+1
ans=0
for j in range(1,int(math.sqrt(k1))+1):
    if k1%j==0:
        if j<=c1 and (k1//j)<=c2:
            ans+=l1[j]*l2[k1//j]
        if j!=k1//j and j<=c2 and (k1//j)<=c1:
            ans+=l2[j]*l1[k1//j]

print(ans)