#include <bits/stdc++.h>
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef __int128 LLL;
typedef std::pair<int, int> pii;
typedef std::pair<LL, int> pli;
typedef std::pair<LL, LL> pll;
typedef unsigned int UI;
typedef std::pair<UI, UI> puu;
#define fi first
#define se second
#define MP std::make_pair
const int MAXN = 1000005, inf = 0x3f3f3f3f, B = 350000;
const LL INF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 998244353;
auto fplus = [](int x, int y){ return x + y >= MOD ? x + y - MOD : x + y; };
auto fminus = [](int x, int y){ return x >= y ? x - y : x + MOD - y; };
auto Fplus = [](int &x, int y){ return x = fplus(x, y); };
auto Fminus = [](int &x, int y){ return x = fminus(x, y); };
template<typename T> T& Fmax(T& x, T y){ return x = x < y ? y : x; };
template<typename T> T& Fmin(T& x, T y){ return x = x < y ? x : y; };
int fpow(int x, LL y = MOD - 2)
{
	int res = 1;
	for (; y; y >>= 1, x = (LL)x * x % MOD)
		if (y & 1) res = (LL)res * x % MOD;
	return res;
}
bool isP[MAXN];
int p[MAXN], cP;
LL n;
void Euler(int n)
{
	memset(isP + 1, true, n);
	isP[1] = false;
	for (int i = 2; i <= n; i++)
	{
		if (isP[i]) p[++cP] = i;
		for (int j = 1; p[j] <= n / i; j++)
		{
			isP[i * p[j]] = false;
			if (i % p[j] == 0) break;
		}
	}
}
LL val[MAXN], f[MAXN];
int id[MAXN], tot , m;
int& ID(LL x){ return x <= B ? id[x] : id[n / x + B]; }
void init_25()
{
	tot = 0;
	for (LL l = 1, r; l <= n; l = r + 1)
		r = n / (n / l), val[ID(n / l) = ++tot] = n / l;
	for (m = 0; p[m + 1] <= n / p[m + 1]; m++) ;
}
void min_25()
{
	init_25();
	for (LL l = 1, r; l <= n; l = r + 1)
		r = n / (n / l), f[ID(n / l)] = n / l;
	for (int i = 1; i <= m; i++)
		for (int j = 1; p[i] <= val[j] / p[i]; j++)
			f[j] -= f[ID(val[j] / p[i])] - i;
}
LL count(LL x){ return f[ID(x)] - f[ID(x / 2)]; }
int calc(LL x)
{
	if (x <= 2) return 1;
	if (x == 3) return 3;
	LL c = count(x);
//	printf("%lld: %lld\n", x, c);
	return (x - c - 1) % MOD * fpow(x % MOD, c) % MOD;
}
int main()
{
	scanf("%lld", &n);
	Euler(B), min_25();
	int ans = 1;
	for (LL l = 1, r; l <= n; l = r + 1)
		r = n / (n / l), ans = (LL)ans * fpow(calc(n / l), r - l + 1) % MOD;
	printf("%d\n", ans);
	return 0;
}