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);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskC {
    final int max = 36000;
    int[] M;
    int a, n;
    String s;
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        a = Integer.parseInt(in.nextLine());
        s = in.nextLine();
        n = s.length();
        long res = 0;
        M = new int[36005];
//        M.put(1, 1);
        for (int i = 0; i < n; ++i) {
            int Sum = 0;
            for (int j = i;  j < n; ++j) {
                Sum += s.charAt(j) - '0';
                M[Sum]++;
            }
        }

        for (int i = 0; i < n; ++i) {
            int Sum = 0;
            for (int j = i;  j < n; ++j) {
                Sum += s.charAt(j) - '0';
                if (Sum != 0 && a % Sum == 0 && a / Sum <= max) {
                    res += M[a / Sum];
                    //out.println(i + " " + j);
                }
                if (Sum == 0 && a == 0)
                    res += n * (n + 1) / 2;
            }
        }

        out.println(res);
    }
}
