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.
//Complete this code or write your own from scratch
import java.util.*;
import java.io.*;
import java.math.*;

public class Solution{
    
    static class Reader
    {
        final private int BUFFER_SIZE = 1 << 25;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
 
        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[1280000]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do
            {
                ret = ret * 10 + c - '0';
            }  while ((c = read()) >= '0' && c <= '9');
 
            if (neg)
                return -ret;
            return ret;
        }
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
    }

    public static void main(String[] argh) throws Exception{
        Reader br = new Reader();
        // BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
        long b= br.nextLong();
        long[] arr= new long[(int)Math.sqrt(b)+1];
        long count= 1;
        int c= 0;
        for(int i=2; i<=Math.sqrt(b); i++){
            if(b%i == 0){
                while(b%i == 0){
                    arr[c] += 1;
                    b /= i;
                }
                c++;
            }
        }
        if(b != 1){
            arr[c]= 1;
            c++;
        }
        for(int i=0; i<c; i++){
            count *= (arr[i]+1);
        }
        System.out.println(count);
    }
}