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>
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF1323B extends PrintWriter {
	CF1323B() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1323B o = new CF1323B(); o.main(); o.flush();
	}

	int[] qu = new int[600];
	void count(int[] aa, int n, int[] kk) {
		for (int i = 0, l = 0; i <= n; i++)
			if (i == n || aa[i] == 0) {
				if (l > 0)
					kk[l]++;
				l = 0;
			} else
				l++;
		int cnt = 0;
		for (int l = 1; l <= n; l++)
			if (kk[l] > 0) {
				qu[cnt++] = l;
				qu[cnt++] = kk[l];
			}
		for (int l_ = 1; l_ <= n; l_++) {
			int k_ = 0;
			for (int h = 0; h < cnt; h += 2) {
				int l = qu[h], k = qu[h + 1];
				if (l_ <= l)
					k_ += (l - l_ + 1) * k;
			}
			kk[l_] = k_;
		}
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		int[] aa = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = sc.nextInt();
		int[] bb = new int[m];
		for (int j = 0; j < m; j++)
			bb[j] = sc.nextInt();
		int[] ka = new int[n + 1];
		count(aa, n, ka);
		int[] kb = new int[m + 1];
		count(bb, m, kb);
		long ans = 0;
		for (int la = 1; la <= k && la <= n; la++)
			if (k % la == 0) {
				int lb = k / la;
				if (lb <= m)
					ans += (long) ka[la] * kb[lb];
			}
		println(ans);
	}
}
