God Water likes to eat meat, fish and chocolate very much, but unfortunately, the doctor tells him that some sequence of eating will make them poisonous.

Every hour, God Water will eat one kind of food among meat, fish and chocolate. If there are 333 continuous hours when he eats only one kind of food, he will be unhappy. Besides, if there are 333 continuous hours when he eats all kinds of those, with chocolate at the middle hour, it will be dangerous. Moreover, if there are 333 continuous hours when he eats meat or fish at the middle hour, with chocolate at other two hours, it will also be dangerous.

Now, you are the doctor. Can you find out how many different kinds of diet that can make God Water happy and safe during NNN hours? Two kinds of diet are considered the same if they share the same kind of food at the same hour. The answer may be very large, so you only need to give out the answer module 100000000710000000071000000007.

Input

The fist line puts an integer TTT that shows the number of test cases. (T≤1000T \le 1000T≤1000)

Each of the next TTT lines contains an integer NNN that shows the number of hours. (1≤N≤10101 \le N \le 10^{10}1≤N≤1010)

Output

For each test case, output a single line containing the answer.

樣例輸入

3
3
4
15

樣例輸出

20
46
435170

題目來源

打表程式碼：

#include<bits/stdc++.h>
using namespace std;
///1  魚
///2  肉
///3  巧克力
int ans;
int judge(int a,int b,int c){
    if((a==b) && (b==c)){
        return 0;
    }
    if(a==3 && c==3 && b!=3 ){
        return 0;
    }
    if(b==3 && a==2&& c==1 )
        return 0;
    if(b==3 && a==1 && c==2 )
        return 0;

    return 1;
}
void dfs(int t,int now,int choose,int last_1,int last_2){
    int haha=judge(last_2,last_1,choose);
    if(haha==0)
        return;
    if(now==t){
        if(last_2!=0&&last_1!=0 &&choose!=0)
        {
          ///  cout<<last_2<<"   "<<last_1<<"    "<<choose<<endl;
            ans++;
        }
        return ;
    }
    ///否則的話就是合法了
    dfs(t,now+1,1,choose,last_1);
    dfs(t,now+1,2,choose,last_1);
    dfs(t,now+1,3,choose,last_1);
}
int main()
{
    int n;
    scanf("%d",&n);
    ans=0;
    int t;
    for(int i=1;i<=n;i++){
        scanf("%d",&t);
        if(t==1)
            ans=3;
        else if(t==2)
            ans=9;
        else{
        ans=0;
        dfs(t,1,1,0,0);
        dfs(t,1,2,0,0);
        dfs(t,1,3,0,0);
        }
        cout<<ans<<endl;
    }
}

然後感覺是線性的，杜教BM板子登場！

程式碼：

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll _,n;
namespace linear_seq{
    const int N=10010;
    ll res[N],base[N],_c[N],_md[N];

    vector<ll> Md;
    void mul(ll *a,ll *b,int k)
    {
        rep(i,0,k+k) _c[i]=0;
        rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (int i=k+k-1;i>=k;i--) if (_c[i])
            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,0,k) a[i]=_c[i];
    }
    int solve(ll n,VI a,VI b)
    {
        ll ans=0,pnt=0;
        int k=SZ(a);
        assert(SZ(a)==SZ(b));
        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
        rep(i,0,k) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (int p=pnt;p>=0;p--)
        {
            mul(res,res,k);
            if ((n>>p)&1)
            {
                for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    VI BM(VI s) {
        VI C(1,1),B(1,1);
        int L=0,m=1,b=1;
        rep(n,0,SZ(s)) {
            ll d=0;
            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n) {
                VI T=C;
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            } else {
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    int gao(VI a,ll n){
        VI c=BM(a);
        c.erase(c.begin());
        rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
      scanf("%lld",&n);
        vector<int>v;
        v.push_back(3);
        v.push_back(9);
        v.push_back(20);
        v.push_back(46);
        v.push_back(106);
        v.push_back(244);
        v.push_back(560);
        v.push_back(1286);
        v.push_back(2956);
        v.push_back(6794);
        printf("%lld\n",linear_seq::gao(v,n-1)%mod);
    }
}