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>
//package com.company;

import java.util.Scanner;

public class CF_1323_B {
    public static int[] res(int a[]) {
        int i=0;
        int res[]=new int[a.length+1];
        while(i<a.length) {
            if(a[i]==0) {
                i++;
                continue;
            }
            int j=i;
            while(i<a.length&&a[i]==1)
                i++;
            for(int len=1;len<=i-j;++len) {
                res[len]+=i-j-len+1;
            }
        }
        return res;
    }
    public static void main(String args[]) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int k=sc.nextInt();
        int a[]=new int[n];
        int b[]=new int[m];
        for(int i=0;i<n;++i) {
            a[i]=sc.nextInt();
        }
        for(int i=0;i<m;++i) {
            b[i]=sc.nextInt();
        }
        int res1[]=res(a);
        int res2[]=res(b);
        long ans=0;
        for(int i=1;i<=a.length;++i) {
            if(k%i==0&&k/i<=m) {
                ans+=res1[i]*1l*res2[k/i];
            }
        }
        System.out.println(ans);

    }

}
