一：題目意思

有，求的值，結果對100000007取模。

二：題目思路

令S(n)為所得結果，則S(n)=3S(n-1)-2S(n-2)+S(n-3)，不會推。用BM模板，它在給定前10來項的情況下，可以求出線性遞推式的任意項，時間複雜度為O(k²log(n))，k為單次匯入的前幾項數量。

原理看不懂（https://zerol.me/2018/02/06/linearly-recurrent-sequence/）。

三：程式碼+BM模板

#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 long long ll;
typedef vector<ll> VI;
typedef pair<ll,ll> PII;
const ll mod=100000007;//注意點 
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;}
 
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];
    }
    ll 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;
    }
    ll 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)));
    }
};
const int MAXN=1e6+5;
ll a[MAXN];
ll qk(ll x,ll y){
    ll ans=1;
    while(y){
        if(y&1) ans=(ans*x)%mod;
        y>>=1;
        x=(x*x)%mod;
    }
    return ans;
}
ll inv(ll x){
    return qk(x,mod-2);
}
ll C(ll n,ll m){
    return a[n]*(inv(a[m]*a[n-m]%mod))%mod;
}
ll T(int x,int y){
    return C(x+y*2+2,x-y);
}
int main() {
    vector<ll>v,vv;
    a[0]=1;
    for(int i=1;i<MAXN;i++) a[i]=(a[i-1]*i)%mod;
    for(int i=0;i<=15;i++){
        int k=i/2;ll t=0;
        for(int j=0;j<=k;j++){
            t+=T(i-j,j);
            t%=mod;
        }
        v.push_back(t);
    }
    int nCase,n;
    scanf("%d", &nCase);
    while(nCase--){
        vv=v;
        scanf("%d",&n);
        printf("%lld\n",linear_seq::gao(vv,n)%mod);
    }
}