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.util.*;
import java.io.*;

public class MainB {

	public static void main(String[] args) throws Exception {
		PrintWriter out = new PrintWriter(System.out);
		Scanner sc = new Scanner(System.in);
		new MainB(sc, out);
		out.close();
	}

	public MainB(Scanner in, PrintWriter out) {
		int n = in.nextInt();
		int m = in.nextInt();
		int k = in.nextInt();

		int i, j;

		int numbers1[] = new int[n + 1];
		int numbers2[] = new int[m + 1];

		long cnt1[] = new long[n + 2];
		long cnt2[] = new long[m + 2];

		for (i = 1; i <= n; i++) {
			numbers1[i] = in.nextInt();
			if (numbers1[i] != 0) {
				numbers1[i] += numbers1[i - 1];
				cnt1[numbers1[i]]++;
			}
		}

		for (i = n; i >= 1; i--) {
			cnt1[i] = cnt1[i] + cnt1[i + 1];
		}

		for (i = 1; i <= m; i++) {
			numbers2[i] = in.nextInt();
			if (numbers2[i] != 0) {
				numbers2[i] += numbers2[i - 1];
				cnt2[numbers2[i]]++;
			}
		}

		for (i = m; i >= 1; i--) {
			cnt2[i] = cnt2[i] + cnt2[i + 1];
		}

		long ans = 0L;

		for (i = 1; i <= n && i <= k; i++) {
			if ((k % i) != 0) {
				continue;
			}
			j = k / i;

			if (j > m) {
				continue;
			}

			ans += cnt1[i] * cnt2[j];
		}

		out.println(ans);

	}
}
