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#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <map>#include <set>#include <cassert>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;}// headint _,n;namespace linear_seq {    const int N=10010;    ll res[N],base[N],_c[N],_md[N];    vector<int> 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) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...//        printf("%d\n",SZ(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() {    for (scanf("%d",&_);_;_--) {        scanf("%d",&n);        printf("%d\n",linear_seq::gao(VI{2,24,96,416,1536,5504,18944,64000,212992,702464},n-1));    }}
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