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;
typedef pair<int,int> pii;
typedef long long int lli;

lli n;

vector< lli > d;

void divisors(lli k)
{
	d.clear();

	for(lli i = 1 ; i*i <= k ; i++)
	{
		if( k%i == 0 )
		{
			if( i != 1 ) d.push_back( i );
			if( i*i != k ) d.push_back( k/i );
		}
	}
}

int main()
{
	scanf("%lld",&n);

	divisors( n - 1 );
	lli ans = d.size();

	divisors( n );

	for(int j = 0 ; j < d.size() ; j++)
	{
		lli curD = d[j];

		lli aux = n;

		while( aux%curD == 0 )
			aux /= curD;

		if( aux%curD == 1 ) ans++;
	}

	printf("%lld\n",ans);
}