https://ac.nowcoder.com/acm/contest/9986/F
牛客新生赛6；

#include <bits/stdc++.h>
using namespace std;
long long n;
long long mod = 998244353;
struct ty
{
    long long x, y;
    //复数形如x+yi
    ty operator* (const ty &b)
    {
        ty ans;
        ans.x = ((x * b.x % mod - y *b.y % mod) % mod + mod % mod);
        ans.y = (x *b.y % mod + y * b.x % mod) % mod;
        return ans;
    }
}a, b;
long long ksm1(long long  a,long long  b)
{
    long long ans = 1, t = a;
    while(b)
    {
        if (b & 1) ans =ans * t% mod;
        t = t * t % mod;
        b>>=1;
    }
    return ans;
}

long long ksm2(ty a,long long b)//此处不反回ty类型因为(1 + i) ^ n + (1 - i) ^ n 虚数都是奇次被抵消了;
{
    ty ans, t = a;
    ans.x = 1;  ans.y =0;
    while(b)
    {
        if (b & 1) ans = ans * t;
        t = t * t;
        b>>=1;
    }
    return ans.x % mod;
}


int main()
{
    a.x = 1, a.y = 1;
    b.x = 1, b.y = mod-1;

    cin >> n;
    long long ans = ((ksm1(2,n)+ ksm2(a, n)) % mod + ksm2(b, n) %mod) * ksm1(4, mod-2)/*乘法逆元*/ %mod;
    cout <<(ans+mod )%mod;

    return 0;

}