#include using namespace std; typedef long long ll; const ll MOD = 998244353; ll pw(ll b, ll e) { ll r=1; b%=MOD; for(;e>0;e>>=1){if(e&1)r=r*b%MOD; b=b*b%MOD;} return r; } const ll inv2=pw(2,MOD-2), inv6=pw(6,MOD-2), inv30=pw(30,MOD-2); ll S2(ll n){n%=MOD;return n*(n+1)%MOD*(2*n+1)%MOD*inv6%MOD;} ll S3(ll n){n%=MOD;ll t=n*(n+1)%MOD*inv2%MOD;return t*t%MOD;} ll S4(ll n){n%=MOD;return n*(n+1)%MOD*(2*n+1)%MOD*((3*n%MOD*n%MOD+3*n-1+MOD)%MOD)%MOD*inv30%MOD;} ll pref(ll m){if(m<=0)return 0;return (2*S4(m)+3*S3(m)+S2(m))%MOD;} ll sum_i_sqrt_i(ll L, ll R) { if(L>R)return 0; ll mL=(ll)sqrtl(L); while((mL+1)*(mL+1)<=L)mL++; while(mL>0&&mL*mL>L)mL--; ll mR=(ll)sqrtl(R); while((mR+1)*(mR+1)<=R)mR++; while(mR>0&&mR*mR>R)mR--; if(mL==mR)return mL%MOD*(L%MOD+R%MOD)%MOD*((R-L+1)%MOD)%MOD*inv2%MOD; ll la=L,lb=(mL+1)*(mL+1)-1; ll lv=mL%MOD*(la%MOD+lb%MOD)%MOD*(lb-la+1)%MOD*inv2%MOD; ll ra=mR*mR,rb=R; ll rv=mR%MOD*(ra%MOD+rb%MOD)%MOD*(rb-ra+1)%MOD*inv2%MOD; ll mv=(pref(mR-1)-pref(mL)+MOD)%MOD; return (lv+mv+rv)%MOD; } ll safe_pow(ll v, int k, ll limit) { ll r=1; for(int i=0;ilimit/v)return limit+1; r*=v; if(r>limit)return limit+1;} return r; } int main(){ ios::sync_with_stdio(false); cin.tie(nullptr); ll N; cin>>N; // Generate change points as (position, k, new_v) triples // When we cross position v^k, the k-th root floor(i^{1/k}) changes from v-1 to v. // But wait: multiple k's can have same position (e.g., 64 = 2^6 = 4^3 = 8^2). // We handle all k's for a given position. // For incremental q update: // q = prod_{k=3}^{K} floor(i^{1/k}) // When we enter block starting at cp = v^k: // old floor(cp^{1/k}) = v-1... no wait. // At cp = v^k: floor(cp^{1/k}) = v (just became v). // Before cp (at cp-1): floor((cp-1)^{1/k}) = v-1. // So the k-th root went from v-1 to v. // Update q: q = q / (v-1) * v (mod MOD). // But q/(...) is modular division: q * inv(v-1) * v. // However, multiple k's can change at the same cp! // We need to handle all of them. // Also, there can be change points where multiple k's change simultaneously. // e.g., at i=64: k=2 (8^2 -> floor changes to 8), k=3 (4^3 -> floor changes to 4), // k=6 (2^6 -> floor changes to 2). // But we're only looking at k>=3, so at 64: k=3 changes (from 3 to 4), k=6 changes (from 1 to 2). // Wait at 64: floor(64^{1/3}) = 4, floor(63^{1/3}) = 3. Yes. // floor(64^{1/6}) = 2, floor(63^{1/6}) = 1. Yes. // floor(64^{1/4}) = floor(2.83) = 2, floor(63^{1/4}) = floor(2.82) = 2. No change. // So at position v^k, the k-th root changes. But also for divisors of k? Not necessarily. // We just need to track (cp, k, v) for each (v, k) pair where v^k <= N. struct CP { ll pos; int k; ll v; }; vector cps_full; int max_k = 1; while((1LL << max_k) <= N) max_k++; for(int k=3; k N) break; cps_full.push_back({vk, k, v}); } } // Sort by position sort(cps_full.begin(), cps_full.end(), [](const CP& a, const CP& b){return a.pos < b.pos;}); // Initial q at i=1: all floor(1^{1/k}) = 1, so q = 1. // Track current value of floor(i^{1/k}) for each k. vector cur_root(max_k + 1, 1); // cur_root[k] = current floor(i^{1/k}) for k>=3 ll cur_q = 1; // = prod_{k>=3} cur_root[k], but all are 1 initially ll total = 0; ll L = 1; int ci = 0; int ncps = cps_full.size(); while(L <= N){ // Find the next change point after L (i.e., smallest cp.pos >= L that we haven't processed) // Actually, let's process: all change points at the same position together. ll R = N; // default: go to end if(ci < ncps) R = min(R, cps_full[ci].pos - 1); // Process block [L, R] if(L <= R){ ll s = sum_i_sqrt_i(L, R); total = (total + cur_q * s) % MOD; } if(ci >= ncps) break; // Advance to next block: process all change points at cps_full[ci].pos ll next_L = cps_full[ci].pos; while(ci < ncps && cps_full[ci].pos == next_L){ int k = cps_full[ci].k; ll v = cps_full[ci].v; ll old_v = cur_root[k]; // should be v-1 // Update q: divide by old_v (if old_v >= 2), multiply by v if(old_v >= 2){ cur_q = cur_q * pw(old_v, MOD-2) % MOD * (v % MOD) % MOD; } else { // old_v = 1 (contributes 1 to product), now becomes v cur_q = cur_q * (v % MOD) % MOD; } cur_root[k] = v; ci++; } L = next_L; } cout << total << "\n"; return 0; }