//#include <atcoder/all>
#include <iostream>
#include <numeric>
#include <cmath>
#include <limits>
#include <stdio.h>
#include <iomanip>
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
using namespace std;
//using namespace atcoder;
using ll = long long;
#define all(A) A.begin(),A.end()
using vll = vector<ll>;
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
using Graph = vector<vector<pair<ll,ll>>>;

vector<ll> power(70);
ll mod = 998244353;
void D(ll A) {
	rep(i, 70) {
		power[i] = 0;
	}
	power[0] = A;
	rep(i, 69) {
		power[i + 1] = power[i] * power[i];
		power[i + 1] %= mod;
	}
}
ll modpow(ll k) {
	ll P = k;
	ll res = 1;
	rep(i, 70) {
		if (P % 2 != 0) {
			res *= power[i];
			res %= mod;
		}
		P /= 2;
	}
	return res;
}//A^kを高速に求める。前処理D(A)、計算ともにO(log INF)
int main() {
	ll N;
	cin >> N;
	ll a = 1;
	D(6);
	a = modpow(N / 2);
	cout << a << endl;
}