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.
//package practice_codes;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.stream.Collectors;

//import codes.Reference.i;

public class LCM {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		long t;
		
		t = i.l();
		HashMap<Long, Integer> map = new HashMap<Long, Integer>();
		for(long i=1;i<=Math.sqrt(t);i++)
		{
			if(t%i==0)
			{
				map.put(t/i, 0);
				map.put(i, 0);
			}
		}
		System.out.println(map.size());
	}
	static private class i
	{
		//input taking methods
		static BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
		
		public static String[] sa()
	    {
	    	try {
				return b.readLine().split(" ");
			} catch (IOException e) {
				
				e.printStackTrace();
			}
	    	return null;
	    }
		
		public static long l()
		{
			try {
				return Long.parseLong(b.readLine());
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
	    	return 0;
		}
		
		public static String s()
		{
			try {
				return b.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
	    	return null;
		}
		
		public static long[] la()
		{
			return Arrays.stream(sa()).parallel().mapToLong(Long::parseLong).toArray();
		}
		
		public static ArrayList<Long> lal()
		{
			return (ArrayList<Long>)Arrays.stream(la()).parallel().boxed().collect(Collectors.toList());
		}
		public static StringBuilder so(String a)
		{
			return new StringBuilder(a);
		}

	}
	/*
	 HashMap<String,Integer> map = new HashMap<String,Integer>();
	 if(!map.containsKey(key))
	 {
	 	map.put(key,value);
	 }
	 else
	 {
	 	map.replace(key,map.get(key));
	 } 
	 */
	static private class algo{
		
		public static boolean isPrime(int a)
		{
			for(int i=2;i<a;i++)
			{
				if(a%i==0)
				{
					return false;
				}
			}
			return true;
		}
		static String reverse(String s)
		{
			return i.so(s).reverse().toString();
		}
		static String replace(String s,int st,int e,String repl)
		{
			return i.so(s).replace(st, e, repl).toString();
		}
		static String insert(String s,int pos,String ins)
		{
			return i.so(s).insert(pos, ins).toString();
		}
		static long sod(long n) 
	    { 
	        long sum; 
	        for (sum = 0; n > 0; sum += n % 10, 
	                                  n /= 10); 
	  
	        return sum; 
	    } 

		static int gcd(int a, int b) 
	    { 
			if (a == 0) 
				return b;  
			return gcd(b % a, a);  
	    } 
		
	    static int lcm(int a, int b) 
	    { 
	        return (a*b)/gcd(a, b); 
	    }
	    static long sumOfSquares(long[] a)
	    {
	    	
	    	return (long) Arrays.stream(a).mapToDouble(i->i*i).sum();
	    }
	    //use of filter
	    //Arrays.stream(a).filter(k->k==1).count();
	    //            array       variable->condition returning boolean  . lot fof functionsa re available
	    
	    static String[] generateBinSeq(int cnt)
	    {
	    	int loopvar=(int) Math.pow(2, cnt);
			String a[][] = new String[cnt+1][loopvar];
			
			for(int i=1;i<=cnt;i++)
			{
				int temp = (int) Math.pow(2, i);
				for(int j=0;j<temp;j++)
				{
					if(i==1 || i==2)
					{
						if(i==1)
						{
							a[i][0]="0";
							a[i][1]="1";
							continue;
						}
						else
						{
							a[2][0]="00";a[2][1]="01";a[2][2]="10";a[2][3]="11";
							continue;
						}
					}
					else
					{
						if(j<(temp/2))
						{
							a[i][j] = a[i-1][j]+"0"; 
						}
						else
						{
							a[i][j] = a[i-1][j%((temp+1)/2)]+"1";
						}
				
					}
						
				}
			}
	    
			return a[cnt];
	    }
	    
	 // USe this snippet for calling
//	    int ndx = algo.bs(a,key,0,a.length-1);
//		if(ndx<0)
//			ndx=0;
//		if(ndx==a.length)
//			ndx=a.length-1;

	    static public int bs(long a[],long key,int s,int e)
		{
			int m = (s+e)/2;
			if(s>e)
				return s;
				
			if(a[m]==key)
				return m;
			else if(a[m]>key)
				return bs(a,key,s,m-1);
			else
				return bs(a,key,m+1,e);
			
		}
	 // USe this snippet for calling
//	    int ndx = bsa(al,key,0,al.size()-1);
//		if(ndx<0)
//			ndx=0;
//		if(ndx==al.size())
//			ndx=al.size()-1;
	    static public int bsa(ArrayList<Long> a,long key,int s,int e)
		{
			int m = (s+e)/2;
			if(s>e)
				return s;
				
			if(a.get(m)==key)
				return m;
			else if(a.get(m)>key)
				return bsa(a,key,s,m-1);
			else
				return bsa(a,key,m+1,e);
			
		}
	    static public long fact(long a)
	    {
	    	long i,fact=1;  
	    	 long number=a;   
	    	  for(i=1;i<=number;i++){    
	    	      fact=fact*i;    
	    	  } 
	    	  return fact;
	    }
	    
	}
}
