Given is a positive integer N.

We will choose an integer K between 2 and N (inclusive), then we will repeat the operation below until N becomes less than K.

* Operation: if K divides N, replace N with N/K; otherwise, replace N with N-K.



In how many choices of K will N become 1 in the end?

Constraints

* 2 \leq N \leq 10^{12}
* N is an integer.

Input

Input is given from Standard Input in the following format:


N


Output

Print the number of choices of K in which N becomes 1 in the end.

Examples

Input

6


Output

3


Input

3141


Output

13


Input

314159265358


Output

9
#include <bits/stdc++.h>
using namespace std;
using ll = long long;


int main() {
    ll N;
    set<ll> S;
    cin >> N;
    if (N == 2) { puts("1"); return 0; }
    S.insert(N);
    for (ll p = 2; p*p <= N; p++) {
        if (N % p != 0) continue;
        ll n = N;
        while (n % p == 0) n /= p;
        if (n % p == 1) S.insert(p);
    }
    N--;
    S.insert(N);
    for (ll p = 2; p*p <= N; p++) {
        if (N % p == 0) {
            S.insert(p);
            S.insert(N/p);
        }
    }
    cout << S.size() << endl;
    return 0;
}
