#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;

using ll=long long;
using ldub=long double;
using lldub=__float128;
using str=string;

using mint=modint;

template<class T=ll>
using tup2=tuple<T,T>;
template<class T=ll>
using tup3=tuple<T,T,T>;
template<class T=ll>
using tup4=tuple<T,T,T,T>;
template<class T=ll>
using tup5=tuple<T,T,T,T,T>;
template<class T=ll>
using tup6=tuple<T,T,T,T,T,T>;
template<class T=ll>
using tup7=tuple<T,T,T,T,T,T,T>;

template<class T=ll>
using vec=vector<T>;
template<class T=ll>
using vec2=vector<vec<T>>;
template<class T=ll>
using vec3=vector<vec2<T>>;
template<class T=ll>
using vec4=vector<vec3<T>>;
template<class T=ll>
using vec5=vector<vec4<T>>;
template<class T=ll>
using vec6=vector<vec5<T>>;

template<class T>
using que=queue<T>;
template<class T>
using Pque=priority_queue<T>;
template<class T>
using pque=priority_queue<T, vector<T>, greater<T>>;

struct Edge{
  ll from,to,w=1,num=-1;
};
using gvec=vector<Edge>;
using gvec2=vector<gvec>;

template<class T>
bool chmax(T &a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>
bool chmin(T &a,T b){if(b<a){a=b;return 1;}return 0;}

#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define ins insert
#define tup make_tuple
#define low lower_bound
#define pops(a) __builtin_popcountll(a)
#define ctz(a) __builtin_ctzll(a)
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define nperm(v) next_permutation(all(v))
#define Yes cout << "Yes" << endl
#define No cout << "No" << endl
#define YN(f) cout << ((f)?"Yes":"No") << endl

const int IINF=2e9;
const ll INF=4e18;

#ifndef LOCAL
#define debug(x)
#endif


/*

Library References:

https://github.com/caz37OwO/library

*/




// =================================================




void solve(){
  ll N=1000010;
  vec<mint> P(N);
  P[0]=1;
  for(ll i=1;i<N;i++) P[i]=P[i-1]*2;
  mint i2=(mint)1/P[1];
  vec2<mint> dp(N,vec<mint>(3));
  dp[0][0]=dp[1][0]=1;
  for(ll i=1;i<N-1;i++){
    dp[i+1][0]+=dp[i][0];
    dp[i+1][1]+=dp[i][0]*(P[i-1]-1);
    dp[i+1][2]+=dp[i][0]*(P[i-1]*(P[i-1]+1)*i2-P[i-1]);
    dp[i+1][2]+=dp[i][1]*(P[i-1]*P[i-1]);
    dp[i+1][2]+=dp[i][2]*(P[i-1]*P[i]);
  }
  ll Q;
  cin >> Q;
  for(ll q=0;q<Q;q++){
    ll n;
    cin >> n;
    cout << (dp[n][0]+dp[n][1]+dp[n][2]).val() << endl;
  }
}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  mint::set_mod(998244353);
  // mint::set_mod(1000000007);

  ll T=1;
  // cin >> T;
  while(T--) solve();
  return 0;
}