import java.util.*;

/**
 * Try each possible f(n) value. For 2<=x<=10^18 we know that n>=2 and f(n)<63.
 * For each f(n) value, find the value m such that m^f(n)=x, if such a value exists.
 * Then check if f(m)=f(n) by counting the divisors of m.
 *
 * @author Finn Lidbetter
 */

public class pd_finn {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long x = sc.nextLong();
        if (x==1) {
            System.out.println(1);
            return;
        }
        long rt = (long)Math.sqrt(x) + 1;
        long best = -1;
        for (int exp=2; exp<=62; exp++) {
            long lo = 2;
            long hi = rt;
            long candidate = -1;
            while (lo<=hi) {
                long mid = lo + (hi - lo) / 2;
                long pow = 1;
                for (int i=0; i<exp; i++) {
                    if (pow<Long.MAX_VALUE / mid) {
                        pow *= mid;
                    } else {
                        pow = x + 1;
                        break;
                    }
                }
                if (pow==x) {
                    candidate = mid;
                    break;
                }
                if (pow<x) {
                    lo = mid + 1;
                } else {
                    hi = mid - 1;
                }
            }
            if (candidate != -1 && (best==-1 || candidate<best)) {
                int divisors = countDivisors((int)candidate);
                if (divisors == exp) {
                    best = candidate;
                }
            }
        }
        System.out.println(best);
    }
    static int countDivisors(int val) {
        int upper = (int)Math.sqrt(val);
        int divisors = 0;
        if (upper*upper==val) {
            divisors = 1;
            upper -= 1;
        }
        for (int i=1; i<=upper; i++) {
            if (val%i==0) {
                divisors += 2;
            }
        }
        return divisors;
    }
}