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.*;
import java.math.BigInteger;

public class codeforcesc {
//
//	static void methods() {
//		Integer arr[] = { 1, 1, 1, 1, 1, 1 };
//		Integer arr2[] = { 1, 1, 1, 1, 1, 1 };
//		List<Integer> l1 = new ArrayList<>();
//		l1 = Arrays.asList(arr);
//		ArrayList<Integer> ll = new ArrayList<>(l1);
//		boolean bb = Arrays.equals(arr, arr2);
//		Collections.reverse(ll);
//		
//	
//		
//	}
//
//	static long gcd(long a, long b) {
//		if (b == 0)
//			return a;
//		return gcd(b, a % b);
//	}
	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();
		}

		TreeMap<Integer, Integer> tm = new TreeMap<>(Collections.reverseOrder());
		TreeMap<Integer, Integer> tm2 = new TreeMap<>(Collections.reverseOrder());
		int count = 0;
		for (int i = 0; i < n; i++) {
			if (a[i] == 1) {
				count++;
				int v = tm.getOrDefault(count, 0);
				tm.put(count, v + 1);
			} else
				count = 0;
		}
		count = 0;
		for (int i = 0; i < m; i++) {
			if (b[i] == 1) {
				count++;
				int v = tm2.getOrDefault(count, 0);
				tm2.put(count, v + 1);
			} else
				count = 0;
		}
		int v = 0;
		for (Map.Entry<Integer, Integer> e : tm.entrySet()) {

			int kk = e.getKey();
			if (kk != tm.firstKey()) {
				tm.put(kk, tm.get(kk) + v);
			}
			v = tm.get(kk);
		}
         v=0;
		for (Map.Entry<Integer, Integer> e : tm2.entrySet()) {

			int kk = e.getKey();
			if (kk != tm2.firstKey()) {
				tm2.put(kk, tm2.get(kk) + v);
			}
			v = tm2.get(kk);
		}
       long ans=0;
		for (Map.Entry<Integer, Integer> e : tm2.entrySet()) {
			int kk = e.getKey();
			long v1=e.getValue();
			long v2=0;
			if(k%kk==0) v2=tm.getOrDefault(k/kk,0);
			//System.out.println("kk="+kk+" k1="+k/kk+" v1="+v1+" v2= "+v2);
			
			ans+=(v1*v2);
		}
         System.out.println(ans);
//		System.out.println(tm.toString());
//		System.out.println(tm2.toString());

	}

}