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.*;

public class Problem626B {
	public static void main(String[] args) throws Exception {
		/*int[] test = new int[100];
		for (int i = 0; i < test.length; i++) {
			if (Math.random() < 0.8) {
				test[i]  = 1;
			}
		}
		System.out.println(Arrays.toString(test));
		System.out.println(Arrays.toString(getCounter(test, 100)));
		System.out.println(Arrays.toString(getCounter2(test, 100)));*/

		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

		StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
		int n = Integer.parseInt(tokenizer.nextToken());
		int m = Integer.parseInt(tokenizer.nextToken());
		int k = Integer.parseInt(tokenizer.nextToken());
		int[] a = new int[n];
		int[] b = new int[m];
		StringTokenizer aT = new StringTokenizer(reader.readLine());
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(aT.nextToken());
		}
		StringTokenizer bT = new StringTokenizer(reader.readLine());
		for (int i = 0; i < m; i++) {
			b[i] = Integer.parseInt(bT.nextToken());
		}
		int[] counterA = getCounter(a, n);
		int[] counterB = getCounter(b, m);
		//System.out.println(Arrays.toString(counterA));
		//System.out.println(Arrays.toString(counterB));
		long sum = 0;
		for (int i = 1; i < counterA.length; i++) {
			if (k % i == 0) {
				int test = k / i;
				if (test < counterB.length) {
					sum += counterA[i] * counterB[test];
				}
			}
		}
		System.out.println(sum);
		reader.close();
	}
	public static int[] getCounter(int[] a, int n) {
		int[] counter = new int[n + 5];
		int numOnes = 0;
		for (int i = 0; i < n; i++) {
			if (a[i] == 1) {
				numOnes++;
			} else {
				counter[numOnes]++;
				numOnes = 0;
			}
		}
		counter[numOnes]++;
		int count = 0;
		int numValues = 0;
		for (int i = n; i > 0; i--) {
			numValues += counter[i];
			count += numValues;
			counter[i] = count;
		}
		numOnes = 0;
		return counter;
	}
	public static int[] getCounter2(int[] a, int n) {
		int[] counter = new int[n + 5];
		int numOnes = 0;
		for (int i = 0; i < n; i++) {
			if (a[i] == 1) {
				numOnes++;
				for (int j = numOnes; j > 0; --j) {
					counter[j]++;
				}
			} else {
				numOnes = 0;
			}
		}
		return counter;
	}
}
