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;
const int maxn = 1e5 + 100;
ll a[maxn];
vector<ll>v;
int main(){
	ll n;
	cin >> n;
	if (n == 2) cout << 1 << endl, exit(0);
	ll x = n - 1;
	int cnt = 1;
	for (int i = 2; i <= sqrt(x); i++){
		if (x%i == 0&&(x/i)>sqrt(n)) cnt++;
	}
	for (int i = 2; i <= sqrt(n); i++){
		x = n;
		while (x%i == 0) x /= i;
		if (x%i == 1) cnt++;
	}
	cout << cnt+1 << endl;
	return 0;
}