Ivan has number b. He is sorting through the numbers a from 1 to 10^{18}, and for every a writes ([a,    b])/(a) on blackboard. Here [a,    b] stands for least common multiple of a and b. Ivan is very lazy, that's why this task bored him soon. But he is interested in how many different numbers he would write on the board if he would finish the task. Help him to find the quantity of different numbers he would write on the board.

Input

The only line contains one integer — b (1 ≤ b ≤ 10^{10}).

Output

Print one number — answer for the problem.

Examples

Input

1


Output

1

Input

2


Output

2

Note

In the first example [a,    1] = a, therefore ([a,    b])/(a) is always equal to 1.

In the second example [a,    2] can be equal to a or 2 ⋅ a depending on parity of a. ([a,    b])/(a) can be equal to 1 and 2.
import java.io.*;
import java.util.*;

public class Solution{
    
    public static void main(String[] args) throws Exception{
        long b = in.readLong();
        boolean[] composite = new boolean[(int)1e5 + 1];
        List<Integer> primes = new ArrayList<>();
        for(int i = 2; i*i < composite.length; i++){
            if(!composite[i]){
                for(int j = i*i; j < composite.length; j+=i){
                    composite[j] = true;
                }
            }
        }
        for(int i = 2; i < composite.length; i++){
            if(!composite[i])
                primes.add(i);
        }
        long count = 1;
        for(int i = 0; i < primes.size() && b > 1; i++){
            int prime = primes.get(i);
            int curr = 0;
            while(b%prime == 0 && b > 1){
                b/=prime;
                curr++;
            }
            if(curr > 0){
                count*=(curr+1);
                // System.out.println(prime+" -> "+ curr +" -> b="+b +" composite919="+composite[919]);
            }
        }
        if(b > 1)   count*=2;
        System.out.println(count);
    }

    static Inputer in;
    static {
        in = new Inputer();
    }

    static class Inputer{
        BufferedReader br;
        Inputer(){
            try{
                br = new BufferedReader(new InputStreamReader(System.in));
            }
            catch(Exception e){}
        }
        public int readInt() throws Exception{
            return Integer.parseInt(readLine());
        }
        public long readLong() throws Exception{
            return Long.parseLong(readLine());
        }
        public int[] readA(String delim) throws Exception{
            String[] s = readLine().split(delim);
            int[] A = new int[s.length];
            for(int i = 0; i < s.length; i++)
                A[i] = Integer.parseInt(s[i]);
            return A;
        }
        public int[] readA() throws Exception{
            String[] s = readLine().split("\\s+");
            int[] A = new int[s.length];
            for(int i = 0; i < s.length; i++)
                A[i] = Integer.parseInt(s[i]);
            return A;
        }
        public long[] readLA() throws Exception{
            String[] s = readLine().split("\\s+");
            long[] A = new long[s.length];
            for(int i = 0; i < s.length; i++)
                A[i] = Long.parseLong(s[i]);
            return A;
        }
        public String readLine() throws Exception{
            return br.readLine();
        }
        public int[] copyA(int[] A){
            int[] B = new int[A.length];
            for(int i= 0 ; i < A.length; i++)
                B[i] = A[i];
            return B;
        }
    }
    static void shuffle(int[] A){
        int n = A.length;
        Random rand = new Random();
        for(int t = 0; t < A.length; t++){
            int i1 = rand.nextInt(n);
            int i2 = rand.nextInt(n);
            int tmp = A[i1];
            A[i1] = A[i2];
            A[i2] = tmp;
        }
    }
}