题目链接：https://nanti.jisuanke.com/t/31721
#include<bits/stdc++.h>
using namespace std;
#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define read(x,y) scanf("%d%d",&x,&y)
#define lrt int l,int r,int rt
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define ll long long
const int maxn =2e5+5;
const int mod=1e9+7;
/*
不要被啥公式吓到，
老老实实的先把
11 21 31
12 22 32
13 23 33
状压成1到9然后推公式，
最后对九个式子用矩阵快速幂即可，
这道题就怕陷入推公式的怪圈。。。。
*/
ll tp[9][9]=///递推公式
{0,0,0,1,0,0,1,0,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,1,0,0,0};
struct jz
{ll a[9][9];jz(){memset(a,0,sizeof(a));}jz operator*( jz y){jz ans;for(int i=0;i<9;i++)for(int j=0;j<9;j++){ans.a[i][j]=0;for(int k=0;k<9;k++)ans.a[i][j]=(ans.a[i][j]+a[i][k]%mod*y.a[k][j]%mod)%mod;}return ans;}
};
jz unit;
ll fpow(ll n)
{jz ans,shu;for(int i=0;i<9;i++) for(int j=0;j<9;j++)shu.a[i][j]=tp[i][j];for(ans=unit;n;n>>=1,shu=shu*shu) if(n&1) ans=shu*ans;///矩阵快速幂ll ret=0;for(int i=0;i<9;i++) for(int j=0;j<9;j++)ret=(ret+ans.a[i][j])%mod;return ret;
}
int main()
{for(int i=0;i<9;i++) unit.a[i][i]=1;int t;scanf("%d",&t);ll n;while(t--){scanf("%lld",&n);if(n==1) puts("3");else if(n==2) puts("9");else printf("%lld\n",fpow(n-2)%mod);}return 0;
}