#include <bits/stdc++.h> 
 
using namespace std;  
 
typedef long long ll;  
typedef pair<int, int> ii;  
 
const int INF = 1e9;  
const ll LINF = 1e18;  
 
// - Đặt f(a, b) = Tổng xor từ a đến b = a ^ (a + 1) ^ (a + 2) ^ ... ^ (b - 1) ^ b
// => f(a, b) = f(0, b) ^ f(0, a - 1)
// => Bài toán đưa về chỉ cần quan tâm đến f(0, n) = Tổng xor từ 0 đến n = 0 ^ 1 ^ 2 ^ ... ^ (n - 1) ^ n
// - Để tính f(0, n) thì ta có nhiều cách như dp digit hay math 
// - Ngoài ra còn có cách đơn giản hơn nhưng phải ra được nhận xét sau:
//   "Tổng xor của 4 số liên tiếp có dạng 4k, 4k + 1, 4k + 2, 4k + 3 (k >= 0) bằng 0" 				
 
// ll f(ll n) {
// 	int len = (n + 1) % 4;  
// 	ll ans = 0;   
// 	for (ll i = n - len + 1; i <= n; i++) {
// 		ans ^= i; 
// 	}
// 	return ans;
// }
 
ll f(ll n) {
	if (n % 4 == 0) return n;   
	if (n % 4 == 1) return 1;  
	if (n % 4 == 2) return n + 1;  
	return 0; 
}
 
int main() {
	ios::sync_with_stdio(false); 
	cin.tie(nullptr); 	
	ll a, b; 
	cin >> a >> b; 
	ll ans = f(b) ^ f(a - 1); 
	cout << ans << '\n'; 
}

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