import java.util.Scanner;
import java.util.Arrays;

class BiggestPrimeNumber {
    /**
        Write a program that finds all prime numbers in the range 1 ... N. Use the Sieve of Eratosthenes algorithm.
        The program should print the biggest prime number which is <= N.

        Input
        On the first line you will receive the number N

        Output
        Print the biggest prime number which is <= N

        SIEVE PSEUDOCODE:
        input: an integer n > 1.
        output: all prime numbers from 2 through n.

        let A be an array of Boolean values, indexed by integers 2 to n, initially all set to true.
        for i = 2, 3, 4, ..., not exceeding √n do
            if A[i] is true
                for j = i2, i2+i, i2+2i, i2+3i, ..., not exceeding n do
                    A[j] := false

        return all i such that A[i] is true.
     **/

    public static int sieveOfEratosthenes(int n) {

        // Initializing the array and setting all values to true:
        boolean[] primes = new boolean[n + 1];
        Arrays.fill(primes, true);

        for (int i = 2; i <= Math.sqrt(n); i++) {
            // If primes[i] is not changed to false, then it is a prime:
            if (primes[i]) {
                // Update all multiples of p to false:
                for (int j = i * i; j <= n; j += i)
                    primes[j] = false;
            }
        }

        // Transfer all primes into a new int array to get the biggest number from for output:
        int[] primesArr = new int[n+1];
        for (int i = 0; i < primes.length; i++) {
            if (primes[i]) {
                primesArr[i] = i;
            }
        }
        Arrays.sort(primesArr);
        return primesArr[primes.length-1];
    }
}

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print(BiggestPrimeNumber.sieveOfEratosthenes(scanner.nextInt()));
    }
}