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.Scanner;

public class Ex20200703_B_CountSubrectangles {

	private static int n[];
	private static int m[];

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int nSize = sc.nextInt();
		int mSize = sc.nextInt();
		int rectangleSize = sc.nextInt();
		n = new int[nSize];
		m = new int[mSize];
		for(int i=0; i<nSize; i++) {
			n[i] = sc.nextInt();
		}
		for(int i=0; i<mSize; i++) {
			m[i] = sc.nextInt();
		}
		sc.close();
		long totalRects = 0;
		for(int i=1; i<=Math.sqrt(rectangleSize); i++) {
			if(rectangleSize % i == 0) {
				totalRects += findRectangles(i, rectangleSize/i);
				if(i != rectangleSize/i) {
					totalRects += findRectangles(rectangleSize/i, i);
				}
			}
		}
		System.out.println(totalRects);
	}
	
	private static long findRectangles(int x, int y) {
		int xCounter = 0;
		int xStart = -1;
		for(int i=0; i<n.length; i++) {
			//System.out.println("\ti=" + i);
			if(n[i] == 0) {
				xStart = i;
				continue;
			}
			if(i-xStart >= x) {
				xCounter++;
				//System.out.println("\tx summing at pos " + i);
			}/*else if(xStart == 0 && i+1 >= x){
				xCounter++;
				//System.out.println("\tx special summing at pos " + i);
			}*/
		}
		int yCounter = 0;
		int yStart = -1;
		for(int i=0; i<m.length; i++) {
			//System.out.println("\ti=" + i);
			if(m[i] == 0) {
				yStart = i;
				continue;
			}
			if(i-yStart >= y) {
				yCounter++;
				//System.out.println("\ty summing at pos " + i);
			}/*else if(yStart == 0 && i+1 >= y){
				yCounter++;
				//System.out.println("\ty special summing at pos " + i);
			}*/
		}
		//System.out.println("Rects with size " + x + "X" + y + ": " + xCounter*yCounter);
		return xCounter * yCounter;
	}
	
	class RectSize{
		int x;
		int y;
	}
}
