You are given an integer number n. The following algorithm is applied to it:

  1. if n = 0, then end algorithm; 
  2. find the smallest prime divisor d of n; 
  3. subtract d from n and go to step 1. 



Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer n (2 ≤ n ≤ 10^{10}).

Output

Print a single integer — the number of subtractions the algorithm will make.

Examples

Input


5


Output


1


Input


4


Output


2

Note

In the first example 5 is the smallest prime divisor, thus it gets subtracted right away to make a 0.

In the second example 2 is the smallest prime divisor at both steps.
import static java.lang.Double.parseDouble;
import static java.lang.Integer.MAX_VALUE;
import static java.lang.Integer.max;
import static java.lang.Long.max;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Math.abs;
import static java.lang.Math.min;
import static java.lang.Math.toRadians;
import static java.lang.System.exit;
import static java.lang.System.gc;

import java.io.*;
import java.util.*;

public class A {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;

    void solve() throws Exception {
        long n = nextLong(), ans = 0;
        int[] pr = sieveEratosthenes(100005);
        for(int x:pr) {
            int cnt = 0;
            while(n % x == 0 && n > 1) {
                cnt++;
                n /= x;
            }
            ans += cnt;
        }
        out.print(ans);
    }

    public static int[] sieveEratosthenes(int n) {
        if (n <= 32) {
            int[] primes = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 };
            for (int i = 0; i < primes.length; i++) {
                if (n < primes[i]) {
                    return Arrays.copyOf(primes, i);
                }
            }
            return primes;
        }

        int u = n + 32;
        double lu = Math.log(u);
        int[] ret = new int[(int) (u / lu + u / lu / lu * 1.5)];
        ret[0] = 2;
        int pos = 1;

        int[] isnp = new int[(n + 1) / 32 / 2 + 1];
        int sup = (n + 1) / 32 / 2 + 1;

        int[] tprimes = { 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 };
        for (int tp : tprimes) {
            ret[pos++] = tp;
            int[] ptn = new int[tp];
            for (int i = (tp - 3) / 2; i < tp << 5; i += tp)
                ptn[i >> 5] |= 1 << (i & 31);
            for (int j = 0; j < sup; j += tp) {
                for (int i = 0; i < tp && i + j < sup; i++) {
                    isnp[j + i] |= ptn[i];
                }
            }
        }

        // 3,5,7
        // 2x+3=n
        int[] magic = { 0, 1, 23, 2, 29, 24, 19, 3, 30, 27, 25, 11, 20, 8, 4, 13, 31, 22, 28, 18, 26, 10, 7, 12, 21, 17,
                9, 6, 16, 5, 15, 14 };
        int h = n / 2;
        for (int i = 0; i < sup; i++) {
            for (int j = ~isnp[i]; j != 0; j &= j - 1) {
                int pp = i << 5 | magic[(j & -j) * 0x076be629 >>> 27];
                int p = 2 * pp + 3;
                if (p > n)
                    break;
                ret[pos++] = p;
                if ((long) p * p > n)
                    continue;
                for (int q = (p * p - 3) / 2; q <= h; q += p)
                    isnp[q >> 5] |= 1 << q;
            }
        }

        return Arrays.copyOf(ret, pos);
    }



    public int lower_bound(long[] a, long v) {
        int low = -1, high = a.length;
        while (high - low > 1) {
            int h = high + low >>> 1;
            if (a[h] >= v) {
                high = h;
            } else {
                low = h;
            }
        }
        return high;
    }


    private int gcd(int a, int b) {
        while (b > 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    private int lcm(int a, int b) {
        return a * (b / gcd(a, b));
    }

    private int[] na(int n) throws IOException {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) a[i] = nextInt();
        return a;
    }

    private long[] nal(int n) throws IOException {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) a[i] = nextLong();
        return a;
    }

    int nextInt() throws IOException {
        return parseInt(next());
    }

    long nextLong() throws IOException {
        return parseLong(next());
    }

    double nextDouble() throws IOException {
        return parseDouble(next());
    }

    String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    public static void main(String[] args) throws Exception {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));
            //long lStartTime = System.currentTimeMillis();
            new A().solve();
            //long lEndTime = System.currentTimeMillis();
            //out.println("Elapsed time in seconds: " + (double)(lEndTime - lStartTime) / 1000.0);
            in.close();
            out.close();
        } catch (Throwable e) {
            e.printStackTrace();
            exit(1);
        }
    }
}