Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
ID: lit20182
TASK: beads
LANG: C++                 
*/
#include<bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
#define mod 1000000007
#define rep(i, a, b) for(ll i = a; i < b; i++)
#define repc(i, a, b, c) for (ll i = a; (c > 0 ? i < b : i > b) and c != 0; i += c)
#define mp make_pair
#define fi first
#define si second
#define pb push_back
#define all(name) (name.begin(), name.end())
#define fact(n) vector<ll> ft(1, 1LL);rep(i, 1, n+1)ft.pb((ft[i-1]*i)%mod);
#define mod_in(a) power(a, mod-2)
#define ncr(n, r) ((ft[n]*mod_in((ft[r]*ft[(n)-(r)])%mod))%mod)
#define deb(x) cout << #x << " " << x << endl;
#define int ll
 
 
ll power(ll x, ll y) {
    ll res = 1;x %= mod;
    while (y > 0) {
        if (y & 1)res = (res*x) % mod;
        y = y>>1;
        x = (x*x) % mod;
    }
    return res%mod;
}
/*_____________________________________________________________________________________________________*/
 
 
signed main()
{
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL);
	int t = 1;
	// cin >> t;
    next :
	while(t--){
        int n, ans;
        cin >> n;
        ans = min(2LL, n-1LL);
        for(int i = 2; i*i <= n; i++){
           int m = n;
           while(m%i == 0)m /= i;
           if(m%i == 1)ans += 1 + ((m==n) and (i*i != n-1));
        }
        cout << ans << endl;
    }
}