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 deque
#import heapq
input=sys.stdin.readline
def solve1(arr,n,i):
    curr=0
    bda=0
    chota=0
    for j in range(n):
        if(arr[j]==1):
            curr+=1
        else:
            bda+=max(0,curr-i+1)
            curr=0
    bda+=max(0,curr-i+1)
    return bda
    
def solve(k,arr,brr):
    n=len(arr)
    m=len(brr)
    i=1
    ans=0
    while(i<=math.sqrt(k)):
        if(k%i==0):
            if(i==k//i):
                bda=solve1(arr,n,i)
                chota=solve1(brr,m,i)
                ans+=bda*chota
            else:
                bda=solve1(arr,n,i)
                chota=solve1(brr,m,k//i)
                ans+=bda*chota
                bda=solve1(arr,n,k//i)
                chota=solve1(brr,m,i)
                ans+=bda*chota
        i+=1
    return ans
                
                
n,m,k=map(int,input().split())
arr=list(map(int ,input().split()))
brr=list(map(int ,input().split()))

ans=solve(k,arr,brr)
print(ans)    


    