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.
//package avito2018;

import java.io.InputStream;
import java.io.PrintStream;
import java.util.Scanner;

public class A {
    public InputStream in = System.in;
    public PrintStream out = System.out;

    private Scanner scan;

    int n;
    int []a;

    void p(Object value) {
        out.print(value);
    }
    void pln(Object value) {
        out.println(value);
    }

    A() {
        scan = new Scanner(in);
    }

    public A redirectIO(InputStream in, PrintStream out) {
        this.in = in;
        this.out = out;
        scan = new Scanner(in);
        return this;
    }

    public static long test(InputStream inBuffer, PrintStream outBuffer){
        A task = new A().redirectIO(inBuffer, outBuffer);
        long from = System.nanoTime();
        task.run();
        return System.nanoTime() - from;
    }

    public static void main(String[] args) {
        new A().run();
    }

    void run(){
        n = scan.nextInt();
        
        if (n == 2)
            p(2);
        else
            p(1);
    }
}

