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 java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.StringTokenizer;

public class Main {
	
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static Scanner sc = new Scanner(System.in);
	static StringTokenizer st;
	
	public static void main(String[] args) {
		
		int n = sc.nextInt();
		int m = sc.nextInt();
		long 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[] lA = gao(a);
		int[] lB = gao(b);
		
		
		long answer = 0;
		int until = Math.max(n, m);
		for(long i = 1 ; i <= until ; i++) {
			if(k % i == 0 && i <= n && k/i <= m) {
				answer += (long)lA[(int) i] * (long)lB[(int) (k/i)];
			}
		}
		
		System.out.println(answer);
		
	}

	private static int[] gao(int[] arr) {
		
		int size = arr.length;
		int ret[] = new int[size+1];
		
		for(int i = 0 ; i < size ; i++) {
			if(arr[i] == 0) {
				continue;
			}
			
			int count = i;
			while(count < size && arr[count] == 1) {
				count++;
			}
			
			int len = count - i;
			
			for(int j = 1 ; j <= len ; j++) {
				ret[j] += len - j + 1;
			}
			i = count;
		}
		
		return ret;
	}

}
