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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for(int i=0; i<(int)(n); i++)
#define REP1(i,a,b) for(int i=a; i<=(int)(b); i++)
#define ALL(x) begin(x),end(x)
#define PB push_back
using namespace std;
typedef int64_t LL;
typedef vector<int> VI;
typedef pair<int,int> PII;

template<class T> inline bool chmax( T &a, const T &b ) { return b>a ? a=b,true : false; }
template<class T> inline bool chmin( T &a, const T &b ) { return b<a ? a=b,true : false; }
template<class T> using MaxHeap = priority_queue<T>;
template<class T> using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template<class T, class F=less<T>> void sort_uniq( vector<T> &v, F f=F() ) {
    sort(begin(v),end(v),f);
    v.resize(unique(begin(v),end(v))-begin(v));
}

LL n;
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;

    LL ans=0;
    vector<LL> v;
    for(LL i=2; i*i<=n; i++) {
    	if ( n%i==0 ) {
    		v.push_back(i);
    		v.push_back(n/i);
    	}
    }

    for(LL i:v) {
    	LL m=n;
    	while ( m%i==0 ) m/=i;
    	if ( m%i==1 ) ans++;
    }

    // for(LL i=2; i*i<=n; i++) {
    // 	if ( n%i==0 ) {
    // 		LL m=n;
    // 		while ( m%i==0 ) m/=i;
    // 		if ( m%i==1 )  {
    // 			ans++;
    // 			cout << i << endl;
    // 		}
    // 		if ( m%(n/m)==1 ) {
    // 			ans++;
    // 			cout << n/m << endl;
    // 		}
    // 	}
    // }
    // cout << ans << '\n';
    n--;
    for(LL i=1; i*i<=n; i++) {
    	if ( n%i==0 ) {
    		ans+=2;
    	}
    }
    // LL ans1=0;
    // for(LL i=2; i<=n+1; i++) {
    // 	LL m=n+1;
    // 	while ( m>=i ) {
    // 		if ( m%i==0 ) m/=i;
    // 		else m-=i;
    // 	}
    // 	if ( m==1 ) {
    // 		cout << i << endl;
    // 		ans1++;
    // 	}
    // }
    // cout << ans << '\n';

    cout << ans << '\n';
    return 0;
}