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>
#define ll long long
#define mp make_pair
#define pb push_back
#define VI vector<int>
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
ll n;
bool f(ll d){
	if(d == 1) return 0;
	ll x = n;
	while(x % d == 0) x /= d;
	return x%d == 1;
}
int main(){
	scanf("%lld", &n);
	int tot = 0;
	for(ll i = 1; i * i <= n; i++){
		if(n%i == 0){
			tot += f(i);
			if(i != n/i) tot += f(n/i);
		}
	}
	n--;
	for(ll i = 1; i * i <= n; i++){
		if(n%i == 0){
			if(i != 1) tot++;
			if(n/i != i) tot++;
		}
	}
	printf("%d\n", tot);
	return 0;
}