n = int(input())
a = 1
for i in range(n,n//2 ,-1):
a = a*i
b = 1
for i in range(1,n//2+1):
b = b*i
c = b//(n//2)
ans = ((a//b)//2)*c*c
print(ans)#include <bits/stdc++.h>
using namespace std;
int MOD = 1e9 + 7;
long long fn(long long n) {
if (n == 1) return 1;
return n * fn(n - 1);
}
int main() {
int n;
cin >> n;
cout << fn(n - 1) / (n / 2);
return 0;
}
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