Ivan has number b. He is sorting through the numbers a from 1 to 10^{18}, and for every a writes ([a,    b])/(a) on blackboard. Here [a,    b] stands for least common multiple of a and b. Ivan is very lazy, that's why this task bored him soon. But he is interested in how many different numbers he would write on the board if he would finish the task. Help him to find the quantity of different numbers he would write on the board.

Input

The only line contains one integer — b (1 ≤ b ≤ 10^{10}).

Output

Print one number — answer for the problem.

Examples

Input

1


Output

1

Input

2


Output

2

Note

In the first example [a,    1] = a, therefore ([a,    b])/(a) is always equal to 1.

In the second example [a,    2] can be equal to a or 2 ⋅ a depending on parity of a. ([a,    b])/(a) can be equal to 1 and 2.


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

/**
 * Copyright © 2018 Chris. All rights reserved.
 *
 * @author Chris
 * 2018/7/9 15:33
 * @see format
 */
public class B {

    private static BufferedReader br;
    private static StreamTokenizer st;
    private static PrintWriter pw;

    static final int INF = 1000000007;
    static final int MOD = 1000000007;

    private static void solve() throws IOException {
        long b = Long.parseLong(nextLine());
        int ans = 0;
        long sqrt = (long) Math.sqrt(b + 0.5);
        for (int i = 1; i <= sqrt; i++) {
            if (b % i == 0) {
                if ((b / i) == i) {
                    ans++;
                } else {
                    ans += 2;
                }
            }
        }
        pw.print(ans);
    }

    static void getDiv(Map<Integer, Integer> map, int n) {
        int sqrt = (int) Math.sqrt(n);
        for (int i = sqrt; i >= 2; i--) {
            if (n % i == 0) {
                getDiv(map, i);
                getDiv(map, n / i);
                return;
            }
        }
        map.put(n, map.getOrDefault(n, 0) + 1);
    }

    public static boolean[] generatePrime(int n) {
        boolean p[] = new boolean[n + 1];
        p[2] = true;

        for (int i = 3; i <= n; i += 2) {
            p[i] = true;
        }

        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (!p[i]) {
                continue;
            }
            for (int j = i * i; j <= n; j += i << 1) {
                p[j] = false;
            }
        }
        return p;
    }

    // 18位素数：154590409516822759
    // 19位素数：2305843009213693951 (梅森素数)
    // 19位素数：4384957924686954497
    static boolean isPrime(long n) { //determines if n is a prime number
        int p[] = {2, 3, 5, 233, 331};
        int pn = p.length;
        long s = 0, t = n - 1;//n - 1 = 2^s * t
        while ((t & 1) == 0) {
            t >>= 1;
            ++s;
        }
        for (int i = 0; i < pn; ++i) {
            if (n == p[i]) {
                return true;
            }
            long pt = pow(p[i], t, n);
            for (int j = 0; j < s; ++j) {
                long cur = llMod(pt, pt, n);
                if (cur == 1 && pt != 1 && pt != n - 1) {
                    return false;
                }
                pt = cur;
            }
            if (pt != 1) {
                return false;
            }
        }
        return true;
    }

    static long llMod(long a, long b, long mod) {
        return (a * b - (long) ((double) a / mod * b + 0.5) * mod + mod) % mod;
//        long r = 0;
//        a %= mod;
//        b %= mod;
//        while (b > 0) {
//            if ((b & 1) == 1) {
//                r = (r + a) % mod;
//            }
//            b >>= 1;
//            a = (a << 1) % mod;
//        }
//        return r;
    }

    static int pow(long a, long n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = (ans * a) % MOD;
            }
            a = (a * a) % MOD;
            n >>= 1;
        }
        return (int) ans;
    }

    static int pow(long a, long n, long mod) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = llMod(ans, a, mod);
            }
            a = llMod(a, a, mod);
            n >>= 1;
        }
        return (int) ans;
    }

    private static long[][] initC(int n) {
        long c[][] = new long[n][n];

        for (int i = 0; i < n; i++) {
            c[i][0] = 1;
        }

        for (int i = 1; i < n; i++) {
            for (int j = 1; j <= i; j++) {
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
            }
        }
        return c;
    }

    /**
     * ps: n >= m, choose m from n;
     */
    private static int c(long n, long m) {
        if (m > n) {
            n ^= m;
            m ^= n;
            n ^= m;
        }
        m = Math.min(m, n - m);

        long top = 1;
        long bot = 1;
        for (long i = n - m + 1; i <= n; i++) {
            top = (top * i) % MOD;
        }
        for (int i = 1; i <= m; i++) {
            bot = (bot * i) % MOD;
        }

        return (int) ((top * pow(bot, MOD - 2)) % MOD);
    }

    static int gcd(int a, int b) {
        if (a < b) {
            a ^= b;
            b ^= a;
            a ^= b;
        }
        while (b != 0) {
            int tmp = a % b;
            a = b;
            b = tmp;
        }
        return a;
    }

    static boolean even(long n) {
        return (n & 1) == 0;
    }

    public static void main(String args[]) throws IOException {
        boolean oj = System.getProperty("ONLINE_JUDGE") != null;
        if (!oj) {
            System.setIn(new FileInputStream("in.txt"));
//            System.setOut(new PrintStream("out.txt"));
        }
        br = new BufferedReader(new InputStreamReader(System.in));
        st = new StreamTokenizer(br);
        pw = new PrintWriter(new OutputStreamWriter(System.out));
        st.ordinaryChar('\''); //指定单引号、双引号和注释符号是普通字符
        st.ordinaryChar('\"');
        st.ordinaryChar('/');

        long t = System.currentTimeMillis();
        solve();
        if (!oj) {
            pw.println("[" + (System.currentTimeMillis() - t) + "ms]");
        }
        pw.flush();
    }

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

    private static String next(int len) throws IOException {
        st.nextToken();
        return st.sval;
    }

    private static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    private static long nextLong() throws IOException {
        return Long.parseLong(nextLine());
    }

    private static double nextDouble() throws IOException {
        st.nextToken();
        return st.nval;
    }

    private static String[] nextSS(String reg) throws IOException {
        return br.readLine().split(reg);
    }

    private static String nextLine() throws IOException {
        return br.readLine();
    }
}
