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 long long ll;
typedef pair<int,int> pi;

ll N, M, Q, A, B, C, X, Y, Z;
string s, t;

vector<ll> v;

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin >> N;
	for (ll i = 2; i*i <= N; ++i){
		ll x = N;
		while (x % i == 0) x /= i;
		x %= i;
		if (x == 1) Z++;
		Y = i;
	}
	//cerr << Y << ' ' << Z << '\n';
	N--;
	for (ll i = 1; i*i < N; ++i){
		if (N%i == 0){
			v.push_back(i);
			v.push_back(N/i);
		}
	}
	for (auto it : v){
		if (it > Y){
			Z++;
		}
	}
	cout << Z+1 << '\n';
}