#include<bits/stdc++.h> using namespace std; #define LL long long const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; long long ans,n; int main() { cin >> n; ans = n % 4; if ( ans == 1 ) cout << 1; else if ( ans == 2 ) cout << n+1; else if ( ans == 3 ) cout << 0; else cout << n; return 0; }

#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define ll long long
#define ld long double
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
using namespace std;
const int N = 1e3 + 10;
const int mod = 1e9 + 7;
ll n;
int main() {
    ll ans = 0;
    cin >> n;
    if(!(n & 1))
        n++, ans = n;
    ll num = (n + 1ll) / 2ll;
    if(num & 1)
        ans ^= 1;
    cout << ans;
    return 0;
}

/*60代码
#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
long long ans = 1;
int main() {
long long n;
cin >> n;
for (int i = 2; i <= n; i++) ans = (ans ^ i);
cout << ans << endl;
}
*/

/*80代码
#include<bits/stdc++.h>
using namespace std;

long long ans = 1;
int main() {
int n;
cin >> n;
while (n != 0) {		
if (n & 1 == 0) {
cout << 1;
return 0;
}
//		cout << n << endl;
n = (n >> 1);
}
cout << 0;
return 0;
}
*/

//100AC
#include<bits/stdc++.h>
using namespace std;

int main() {
long long n;
cin >> n;
int k = n % 4;
if (k == 1) cout << 1;
else if (k == 2) cout << n + 1;
else if (k == 3) cout << 0;
else cout << n;
return 0;
}

/*找规律:
1. 1
2. 3
3. 0
4 .4

5. 1
6. 7
7 .0
8. 9
...
*/
#include<iostream>
using namespace std;
long long n;
signed main(){
    cin>>n;
    switch (n%4)
    {
    case 0:
        cout<<n;
        break;
    case 1:
        cout<<1;
        break;
    case 2:
        cout<<n+1;
        break;
    default:
        cout<<0;
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define int long long
int n;
signed main(){
	cin>>n;
	if(n%4==3) cout<<0<<endl;
	if(n%4==0) cout<<n<<endl;
	if(n%4==1) cout<<1<<endl;
	if(n%4==2) cout<<((n/4+1)*4)-1<<endl;
    return 0;
}