$$\begin{gathered} \sum _{d\mid n}f(d){\sigma }_p(\frac{n}{d})={\sigma }_q(n) \\ f\ast {\sigma }_p={\sigma }_q \\ 有{\sigma }_k=\sum _{d\mid n}{d}^k=i{d}_k\ast I \\ f\ast i{d}_p\ast I=i{d}_q\ast I \\ \sum _{d\mid n}\mu (d)=\mu \ast I \\ 对上面式子同时卷上一个\mu \\ f\ast i{d}_p=i{d}_q \\ 因为i{d}_k是一个完全积性函数，所以i{d}_p^{-1}=\mu \times i{d}_p \\ i{d}_k\ast (\mu \times i{d}_p)=\sum _{d\mid n}{d}^k\times \mu (\frac{n}{d})\times (\frac{n}{d}{)}^k=\sum _{d\mid n}\mu (d)=ϵ \\ f=i{d}_q\ast (\mu \times i{d}_p) \\ f(n)=\sum _{d\mid n}\mu (d)\times d^p\times (\frac{n}{d}{)}^q \\ f(1)=1 \\ f(n)=n^q-n^p(n\in primes) \\ f(ab)=f(a)\times f(b)(gcd(a,b)=1) \\ f(n^x)=n^q-n^p{n}^{(x-1)q}=n^{xq}-n^{p+(x-1)q}(n\in primes) \\ 然后就可以线性筛了 \end{gathered}$$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e7 + 10, mod = 998244353;ll prime[N], minnp[N], f[N], cnt, p, q, n;bool st[N];ll quick_pow(ll a, int n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans;
}void init() {f[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[++cnt] = i;f[i] = (quick_pow(i, q) - quick_pow(i, p) + mod) % mod;minnp[i] = 1;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {minnp[i * prime[j]] = minnp[i] + 1;f[i * prime[j]] = f[i / quick_pow(prime[j], minnp[i])] * (quick_pow(prime[j], q * (minnp[i * prime[j]])) - quick_pow(prime[j], p + minnp[i] * q) + mod) % mod;break;}minnp[i * prime[j]] = 1;f[i * prime[j]] = f[i] * f[prime[j]] % mod;}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cin >> n >> p >> q;init();ll ans = 0;for(int i = 1; i <= n; i++) {ans ^= f[i];}cout << ans << "\n";return 0;
}