结论：对于n，答案为$F_{2^n}$

呃，其实找规律也能找出来的

#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
long long t;
long long n[1001];
long long nm;
long long f[1000001],c[1000001];
const long long mod=1000000007;
//const double modc=1000000007;
//long long c;
long long x;
long long d;
long long ans;
int main(){
    scanf("%lld",&t);
    for(long long i=1;i<=t;++i){
        scanf("%lld",&n[i]);
        nm=max(nm,n[i]);
    }
    f[0]=f[1]=1;
    c[0]=1;
    c[1]=1;
    for(long long i=2;i<=2*nm;++i){
        f[i]=f[i-1]+f[i-2];
        f[i]%=mod;
    }
    for(long long i=1;i<=t;++i){
        cout<<f[2*n[i]]<<endl;
    }
    return 0;
}
/*
*/