Chouti was doing a competitive programming competition. However, after having all the problems accepted, he got bored and decided to invent some small games.

He came up with the following game. The player has a positive integer n. Initially the value of n equals to v and the player is able to do the following operation as many times as the player want (possibly zero): choose a positive integer x that x<n and x is not a divisor of n, then subtract x from n. The goal of the player is to minimize the value of n in the end.

Soon, Chouti found the game trivial. Can you also beat the game?

Input

The input contains only one integer in the first line: v (1 ≤ v ≤ 10^9), the initial value of n.

Output

Output a single integer, the minimum value of n the player can get.

Examples

Input


8


Output


1


Input


1


Output


1

Note

In the first example, the player can choose x=3 in the first turn, then n becomes 5. He can then choose x=4 in the second turn to get n=1 as the result. There are other ways to get this minimum. However, for example, he cannot choose x=2 in the first turn because 2 is a divisor of 8.

In the second example, since n=1 initially, the player can do nothing.
import java.util.*;
import java.lang.Math;

public class test
{
    public static void main(String args[])
    {
        Scanner s=new Scanner(System.in);
        int n=s.nextInt();
        if(n==2)
        System.out.println(2);
        else
        System.out.println(1);
        /*long[] a=new long[n];
        for(int i=0;i<n;i++)
        {
            a[i]=n-s.nextLong();
        }
        Arrays.sort(a);
        long sum=0;
        for(int i=1;i<n;i++)
        {
            if(a[i]!=a[i-1])
            {
                sum+=a[i-1];
            }
        }
        sum+=a[n-1];
        if(sum==n)
        {
            System.out.println("Possible");
            for(int i=0;i<n;i++)
                System.out.print(a[i]+" ");
        }
        else
            System.out.println("Impossible");*/
    }
}