BM-找规律

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (ll i=a;i<n;i++)
#define per(i,a,n) for (ll 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) ((ll)(x).size())
typedef long long ll;
typedef vector<ll> VI;
typedef pair<ll, ll> 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 ll N = 10010;
    ll res[N], base[N], _c[N], _md[N];
    vector<ll> Md;
    void mul(ll *a, ll *b, ll 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 (ll 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;
    ll 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 (ll p = pnt; p >= 0; p--) 
    {
        mul(res, res, k);
        if ((n >> p) & 1) 
        {
            for (ll 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);
    ll 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)));
}
};
int main() 
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld", &n);
        VI vec;
        vec.push_back(3);
        vec.push_back(9);
        vec.push_back(20);
        vec.push_back(46);
        vec.push_back(106);
        vec.push_back(244);
        vec.push_back(560);
        vec.push_back(1286);
        vec.push_back(2956);
        vec.push_back(6794);
        vec.push_back(15610);
        //printf("dqwqd\n");
        printf("%lld\n", linear_seq::gao(vec, n - 1)%mod);
    }
}