You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.

A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.

Input

The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).

Output

Print a single integer — the answer to a problem.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

10
12345


Output

6


Input

16
439873893693495623498263984765


Output

40
import java.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskA {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int a=in.nextInt();
        String s=in.next();
        int n=s.length();
        int[] tr=new int[40000];
        for (int i = 0; i < n; i++) {
            int sum=0;
            for (int j = i; j < n; j++) {
                sum+=s.charAt(j)-'0';
                tr[sum]++;
            }
        }
        if (a==0) {
            out.println( (long) tr[0]*(n*(n+1)-tr[0]) );

        }else{
            long ans=0;
            for (int i = 1; (long)i*i <= a; i++)if (a%i==0) {
                if (a/i>=40000) continue;
                ans+=(long)tr[i]*tr[a/i];
                if (a/i!=i)
                    ans+=(long)tr[i]*tr[a/i];
            }
            out.println(ans);
        }
    }
}

