思路:矩阵快速幂。推一下初始矩阵就好了
#include
#define LL long long
#define fi first
#define se second
#define mp make_pair
#define pb push_back
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
const int N = 2e5 +11;
const LL mod=123456789;
LL A[6][6]={
{0,2,0,0,0,0},
{1,1,0,0,0,0},
{0,1,1,0,0,0},
{0,0,3,1,0,0},
{0,0,3,2,1,0},
{0,0,1,1,1,1}
};
LL P[6];
struct uzi
{
LL a[6][6];
void res(){
memset(a,0,sizeof a);
}
};
int cnt;
uzi operator *(const uzi &x,const uzi &y){
uzi ans;
ans.res();
for(int i=0;i<6;i++){
for(int j=0;j<6;j++){
LL mid=0;
for(int k=0;k<6;k++){
mid+=x.a[i][k]*y.a[k][j]%mod;
mid%=mod;
}
ans.a[i][j]=mid;
}
}
return ans;
}
LL POW(LL n){
uzi ans,mid;
ans.res();
mid.res();
for(int i=0;i<6;i++)ans.a[i][i]=1;
for(int i=0;i<6;i++)for(int j=0;j<6;j++)mid.a[i][j]=A[i][j];
while(n){
if(n%2){
ans=mid*ans;
}
mid=mid*mid;
n/=2;
}
LL sum=0;
for(int i=0;i<6;i++){
sum=sum+P[i]*ans.a[i][0]%mod;
sum%=mod;
}
return sum;
}
int main(){
ios::sync_with_stdio(false);
P[0]=1,P[1]=2,P[2]=27,P[3]=9,P[4]=3,P[5]=1;
int t;
for(cin>>t;t;t--){
LL n;
cin>>n;
cout<<POW(n-1)<<endl;
}
return 0;
}