#include using namespace std; #include using namespace atcoder; template inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); } template inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); } #define rep(i, n) for (long long i = 0; i < (long long)(n); i++) #define rep2(i, m ,n) for (int i = (m); i < (long long)(n); i++) #define REP(i, n) for (long long i = 1; i < (long long)(n); i++) typedef long long ll; #define updiv(N,X) (N + X - 1) / X #define l(n) n.begin(),n.end() #define YesNo(Q) Q==1?cout<<"Yes":cout<<"No" using P = pair; using mint = modint; const int MOD = 998244353LL; const ll INF = 999999999999LL; vector fact, fact_inv, inv; /* init_nCk :二項係数のための前処理 計算量:O(n) */ template void input(vector &v){ rep(i,v.size()){cin>>v[i];} return; } void init_nCk(int SIZE) { fact.resize(SIZE + 5); fact_inv.resize(SIZE + 5); inv.resize(SIZE + 5); fact[0] = fact[1] = 1; fact_inv[0] = fact_inv[1] = 1; inv[1] = 1; for (int i = 2; i < SIZE + 5; i++) { fact[i] = fact[i - 1] * i % MOD; inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD; fact_inv[i] = fact_inv[i - 1] * inv[i] % MOD; } } /* nCk :MODでの二項係数を求める(前処理 int_nCk が必要) 計算量:O(1) */ long long nCk(int n, int k) { assert(!(n < k)); assert(!(n < 0 || k < 0)); return fact[n] * (fact_inv[k] * fact_inv[n - k] % MOD) % MOD; } long long modpow(long long a, long long n, long long mod) { long long res = 1; while (n > 0) { if (n & 1) res = res * a % mod; a = a * a % mod; n >>= 1; } return res; } ll POW(ll a,ll n){ long long res = 1; while (n > 0) { if (n & 1) res = res * a; a = a * a; n >>= 1; } return res; } ll grundy(ll a){ if(a<=2){return 1;} else if(a<=4){return 2;} else if(a%3==1){return 4*((a-2)/3);} else{return 4*((a-4)/3);} } void solve(){ int n;cin>>n; vector v(n); rep(i,n){cin>>v[i];} ll prod = 0; rep(i,n){ prod ^= grundy(v[i]); } if(prod==0){cout<<"Bob"<>t; mint a4 = 249561089; mint a2 = 499122177; rep(test,t){ ll a;cin>>a; ll ret = 0; if(a%2==0){ret = ((a/2%998244352LL)*((a-1)%998244352LL));} else{ret = ((a%998244352LL)*((a-1)/2%998244352LL));} mint ans = ((1-(a2*modpow(249561089,a-2,998244353)))*(a-1)+1)*modpow(2,ret,998244353LL); cout << ans.val() << endl; } } /*MOD=998244353 T=int(input()) one_half=pow(2,-1,MOD) three_quarters=pow(4,-1,MOD)*3 for _ in range(T): N=int(input()) print((1-(one_half*pow(three_quarters,N-2,MOD)))*(N-1)+1)*pow(2,N*(N-1)/2,MOD);*/