模板：往里面扔数就可以了（据说>=8个？）%%%dls
https://www.cnblogs.com/zzqsblog/p/6877339.html
#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(); //至少8项，越多越好。 printf("%lld\n",linear_seq::gao(v,n-1)%mod);}
}
